0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric...
Transcript of 0 2 2 2 2€¦ · Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric...
56 56
x y 1
sin x 2
coordinates
0 y 1
sin 0
= − 2 1
⋅ 0 = 0 = −
2
(0, 0)
π
2
1 π y = − sin
2 2 1
⋅1 = − 1 = −
2 2
π 1 , − 2 2
π y 1
sin π = −
2 1
⋅ 0 = 0 = −
2
(π , 0)
3π
2
1 3π y = − sin
2 2 1
(−1) = 1 = −
2 2
3π 1 , 2 2
2π y 1
sin 2π = −
2 1
= − ⋅ 0 = 0 2
(2π , 0)
x y = 3sin x coordinates
0 y = 3sin 0 = 3 ⋅ 0 = 0 (0, 0)
π
2 y = 3sin
π = 3 ⋅1 = 3
2
π , 3
2
π y = 3sin x = 3 ⋅ 0 = 0 (π , 0)
3π
2 y = 3sin
3π 2
= 3(−1) = −3
3π , −3
2
2π y = 3sin 2π = 3 ⋅ 0 = 0 (2π , 0)
x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π
adding quarter- periods. The five x-values are
x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
2 2
x = π + π
= 3π x = π +
π =
3π 2 2
2 2
x = 3π
+ π
= 2π 2 2
Evaluate the function at each value of x.
x = 3π
+ π
= 2π 2 2
Evaluate the function at each value of x.
= −
Connect the five points with a smooth curve and
graph one complete cycle of the given function with
the graph of y = sin x .
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Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions
3π 1 y = 2sin ⋅ 3π
32 π
= 2sin 2
= 2 ⋅ (−1) = −2
(3π , − 2)
4π y = 2sin 1
⋅ 4π 2
= 2sin 2π = 2 ⋅ 0 = 0
(4π , 0)
x y = 2sin 1
x 2
coordinates
0 y = 2sin 1
⋅ 0
2
= 2sin 0 = 2 ⋅ 0 = 0
(0, 0)
π y = 2sin 1
⋅π
= 2sin π
= 2 ⋅1 = 2 2
(π , 2)
2π y = 2sin 1
⋅ 2π
= 2 sin π = 2 ⋅ 0 = 0
(2π , 0)
3
Connect the five key points with a smooth curve and
graph one complete cycle of the given function with
the graph of y = sin x . Extend the pattern of each
graph to the left and right as desired. Connect the five key points with a smooth curve and
graph one complete cycle of the given function. Extend
the pattern of the graph another full period to the right.
3. The equation y = 2sin 1
x is of the form 2
y = A sin Bx with A = 2 and B = 1
. 2
The amplitude is A = 2 = 2 .
The period is 2π =
2π = 4π .
π B 1 4. The equation
2 y = 3sin 2 x −
is of the form
Find the x–values for the five key points by dividing
the period, 4π , by 4, period
= 4π
= π , then by y = A sin(Bx − C) with A = 3, B = 2, and C =
π . The
34 4
adding quarter-periods. amplitude is A = 3 = 3 .
The five x-values are The period is 2π
= 2π
= π .x = 0 B 2 x = 0 + π = π πx = π + π = 2π The phase shift is
C = 3 =
π ⋅
1 = π
.x = 2π + π = 3π B 2 3 2 6x = 3π + π = 4π Find the x-values for the five key points by dividingEvaluate the function at each value of x.
the period, π , by 4, period
= π
, then by adding4 4
quarter-periods to the value of x where the cycle
begins, x = π
. 6
The five x-values are
x = π 6 π π 2π 3π 5π
2
x = + = + = 6 4 12 12 12
x = 5π
+ π
= 5π
+ 3π
= 8π
= 2π
12 4 12 12 12 3
x = 2π
+ π
= 8π
+ 3π
= 11π
3 4 12 12 12 2
11π π
11π 3π
14π 7π
x = + = + = = 12 4 12 12 12 6
Evaluate the function at each value of x.
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Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions
x y = 3sin 2 x −
π
coordinates
π
6 y = 3sin
2 ⋅ π
− π
= 3sin 0 = 3 ⋅ 0 = 0
π , 0
5π
12 y = 3sin
2 ⋅
5π − π
= 3sin 3π
= 3sin π
6 2 = 3 ⋅1 = 3
5π , 3
2π
3 y = 3sin
2 ⋅
2π − π
= 3sin 3π
= 3sin π 3
= 3 ⋅ 0 = 0
2π , 0
11π
12 y = 3sin
2 ⋅
11π − π
= 3sin 9π
= 3sin 3π
6 2 = 3 (−1) = −3
11π , − 3
7π
6 y = 3sin
2 ⋅
7π − π
= 3sin 6π
= 3sin 2π 3
= 3 ⋅ 0 = 0
6
x y = −4 cos π x coordinates
0 y = −4 cos (π ⋅ 0) = −4 cos 0 = −4
(0, –4)
1
2
1 y = −4 cos π ⋅
2
= −4 cos π
= 0 2
1 2
, 0
1 y = −4 cos(π ⋅1) = −4 cos π = 4
(1, 4)
3
2 y = −4 cos
π ⋅
3
3π
= −4 cos = 0 2
3 , 0
2 y = −4 cos(π ⋅ 2) = −4 cos 2π = −4
(2, –4)
x = 0
1 1 3
6 3
12 3
6
12
x = 0 + = 2 2
x = 1
+ 1
= 1 2 2
x = 1 + 1
= 3
2 2
x = 3
+ 1
= 2 2 2
Evaluate the function at each value of x.
3 3
3
12 3
12
6 3
7π , 0
2
2
Connect the five key points with a smooth curve
and graph one complete cycle of the given graph.
Connect the five key points with a smooth curve and
graph one complete cycle of the given function. Extend
the pattern of the graph another full period to the left.
5. The equation
y = −4 cos π x is of the form
6. y = 3
cos(2 x + π ) = 3
cos(2 x − (−π )) 2 2
y = A cos Bx with A = −4, and B = π . The equation is of the form y = A cos(Bx − C ) with
Thus, the amplitude is A = −4 = 4 . A = 3
, B = 2 , and C = −π .
The period is 2π
= 2π
= 2 . 2
B π Thus, the amplitude is A = 3 =
3 .
Find the x-values for the five key points by dividing
the period, 2, by 4, period
= 2
= 1
, then by adding
2 2
The period is 2π
= 2π
= π .4 4 2 B 2
quarter periods to the value of x where the cycle
begins. The five x-values are The phase shift is
C =
−π = − π
. B 2 2
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Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions
x y = 3
cos(2x + π ) 2
coordinates
− π 2
y = 3
cos(−π + π ) 2
= 3
⋅1 = 3
2 2
− π
, 3
− π 4
y = 3
cos − π
+ π 2
2
= 3
⋅ 0 = 0 2
− π
, 0
0 y =
3 cos(0 + π )
2
= 3
⋅ −1 = − 3
2 2
0, −
3
2
π
4 y =
3 cos
π + π
2 2
= 3
⋅ 0 = 0 2
π , 0
4
π
2
3 y = cos(π + π )
2
= 3
⋅1 = 3
2 2
π 3 , 2 2
x y = 2 cos x + 1 coordinates
0 y = 2 cos 0 + 1 = 2 ⋅1 + 1 = 3
(0, 3)
π
2
π y = 2 cos + 1
2 = 2 ⋅ 0 + 1 = 1
π 2
π y = 2 cos π + 1 = 2 ⋅ (−1) + 1 = −1
(π , − 1)
3π
2
3π y = 2 cos + 1
= 2 ⋅ 0 + 2
1
1 =
3π , 1 2
2π y = 2 cos 2π + 1 = 2 ⋅1 + 1 = 3
(2π , 3)
= −
Find the x-values for the five key points by dividing
the period, π , by 4, period
= π
, then by adding 4 4
quarter-periods to the value of x where the cycle
begins, x π
. 2
The five x-values are
π x = −
2 π π π
x = − + = − 2 4 4
7. The graph of
y = 2 cos x + 1 is the graph ofπ π
x = − + = 0 4 4
x = 0 + π
= π
y = 2 cos x shifted one unit upwards. The period for
both functions is 2π . The quarter-period is
4 4 2π or
π . The cycle begins at x = 0. Add quarter-
x = π
+ π
= π
4 4 2 Evaluate the function at each value of x.
4 2 periods to generate x-values for the key points.
x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
2 2
x = π + π
= 3π
4
2 2
x = 3π
+ π
= 2π 2 2
Evaluate the function at each value of x.
, 1
Connect the five key points with a smooth curve and
graph one complete cycle of the given graph.
By connecting the points with a smooth curve, we
obtain one period of the graph.
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Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions
6 2
8. Select several values of x over the interval.
x
0 π
4
π
2
3π
4 π
5π
4
3π
2
7π
4 2π
y1 = 2 sin x 0 1.4 2 1.4 0 −1.4 −2 −1.4 0
y2 = cos x 1 0.7 0 −0.7 −1 −0.7 0 0.7 1
y = 2 sin x + cos x 1 2.1 2 0.7 −1 −2.1 −2 −0.7 1
9. A, the amplitude, is the maximum value of y. The graph shows that this maximum value is 4, Thus,
π , and period =
2π .
A = 4 . The period is
2
Thus,
B
π =
2π
2 Bπ B = 4π
B = 4
Substitute these values into
y = A sin Bx . The graph is modeled by
y = 4sin 4 x .
10. Because the hours of daylight ranges from a minimum of 10 hours to a maximum of 14 hours, the curve oscillates about
the middle value, 12 hours. Thus, D = 12. The maximum number of hours is 2 hours above 12 hours. Thus, A = 2. The
graph shows that one complete cycle occurs in 12–0, or 12 months. The period is 12.
Thus, 12 = 2π B
12B = 2π
B = 2π
= π
12 6
The graph shows that the starting point of the cycle is shifted from 0 to 3. The phase shift, C
, is 3. B
3 = C
B
3 = C
π 6
π = C
2
Substitute these values into y = A sin(Bx − C ) + D . The number of hours of daylight is modeled by
y = 2sin π
x − π
+ 12 .
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Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions
3π
2
3π y = 4sin
2 = 4(−1) = −4
3π , − 4 2
2π y = 4sin 2π = 4 ⋅ 0 = 0 (2π , 0)
x y = 5sin x coordinates
0 y = 5sin 0 = 5 ⋅ 0 = 0 (0, 0)
π
2
π y = 5sin
2 = 5 ⋅1 = 5
π 2
, 5
π y = 5sin π = 5 ⋅ 0 = 0 (π , 0)
3π
2 y = 5sin
3π = 5(−1) = −5
2
3π , − 5
2π y = 5sin 2π = 5 ⋅ 0 = 0 (2π , 0)
x y = 4sin x coordinates
0 y = 4sin 0 = 4 ⋅ 0 = 0 (0, 0)
π
2 y = 4sin
π = 4 ⋅1 = 4
2
π , 4
2
π y = 4sin π = 4 ⋅ 0 = 0 (π , 0)
Concept and Vocabulary Check 2.1
1. A ; 2π B
2. 3; 4π
3. π ; 0; π
;
π ;
3π ; π
Connect the five key points with a smooth curve and
graph one complete cycle of the given function with4 2 4 the graph of y = sin x .
4. C
; right; left B
5. A ; 2π B
6. 1
; 2π
2 3
7. false
8. true
2. The equation
y = 5sin x is of the form
y = A sin x
9. true with A = 5. Thus, the amplitude is A = 5 = 5 .
10. true The period is 2π . The quarter-period is 2π or
π .
Exercise Set 2.1
4 2 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points.
x = 0
1. The equation
y = 4sin x
is of the form
y = A sin x x = 0 + π
= π
2 2with A = 4. Thus, the amplitude is A = 4 = 4 . x =
π + π
= π 2 2
The period is 2π . The quarter-period is 2π or
π . x = π +
π =
3π4 2
The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points.
x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
x = π + π
= 3π
2 2
x = 3π
+ π
= 2π 2 2
Evaluate the function at each value of x.
2 2
x = 3π
+ π
= 2π 2 2
Evaluate the function at each value of x.
2
Connect the five key points with a smooth curve and
graph one complete cycle of the given function with
the graph of y = sin x .
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62 62
Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions
x y = 1
sin x 4
coordinates
0 y = 1
sin 0 = 1
⋅ 0 = 0 4 4
(0, 0)
π
2 y =
1 sin π
= 1
⋅1 = 1
4 2 4 4
π ,
1
π y = 1
sin π = 1
⋅ 0 = 0 4 4
(π , 0)
3π
2 y =
1 sin
3π =
1 (−1) = −
1 4 2 4 4
3π , −
1 2 4
2π 1 1
y = sin 2π = ⋅ 0 = 0 4 4
(2π , 0)
x y =
1 sin x
3
coordinates
0 y =
1 sin 0 =
1 ⋅ 0 = 0
3 3
(0, 0)
π
2 y =
1 sin π
= 1
⋅1 = 1
3 2 3 3
π ,
1 2 3
π y =
1 sin π =
1 ⋅ 0 = 0
3 3
(π , 0)
3π
2 y =
1 sin
3π 3 2
= 1
(−1) = − 1
3 3
3π , −
1
2 3
2π y =
1 sin 2π =
1 ⋅ 0 = 0
3 3
(2π , 0)
3. The equation
y = 1
sin x is of the form 3
y = A sin x
4. The equation
1
y = 1
sin x is of the form 4
y = A sin x
1 1
with A = 1
. Thus, the amplitude is A = 1 =
1 .
with A = . Thus, the amplitude is 4
A = = . 4 4
3 3 3 2π π
The period is 2π . The quarter-period is 2π
or π
. The period is 2π . The quarter-period is or .
4 24 2 The cycle begins at x = 0. Add quarter-periods to
The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points.
x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π
generate x-values for the key points. x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
2 2
x = π + π
= 3π
x = π + π
= 3π
2 22 2
x = 3π
+ π
= 2π 2 2
Evaluate the function at each value of x.
x = 3π
+ π
= 2π 2 2
Evaluate the function at each value of x.
2 4
Connect the five key points with a smooth curve and
graph one complete cycle of the given function withConnect the five key points with a smooth curve and graph one complete cycle of the given function with
the graph of y = sin x .
the graph of y = sin x .
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63 63
Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions
x y = −4sin x coordinates
0 y = −4sin 0 = −4 ⋅ 0 = 0 (0, 0)
π
2 y = −4sin
π = −4 ⋅1 = −4
2
π , − 4
π y = −4sin π = −4 ⋅ 0 = 0 (π , 0)
3π
2
3π y = −4 sin = −4(−1) = 4
2
3π , 4 2
2π y = −4sin 2π = −4 ⋅ 0 = 0 (2π , 0)
x y = −3sin x coordinates
0 y = −3sin x = −3 ⋅ 0 = 0
(0, 0)
π
2 y = −3sin
π 2
= −3 ⋅1 = −3
π , − 3
π y = −3sin π = −3 ⋅ 0 = 0
(π , 0)
3π
2 y = −3sin
3π 2
= −3(−1) = 3
3π , 3
2
2π y = −3sin 2π = −3 ⋅ 0 = 0
(2π , 0)
6. The equation y = −4sin x is of the form y = A sin x
with A = –4. Thus, the amplitude is A = −4 = 4 .
The period is 2π . The quarter-period is 2π or
π .
4 2 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x = 0
x = 0 + π
= π
5. The equation y = −3sin x is of the form y = A sin x 2 2 π π
with A = –3. Thus, the amplitude is A = −3 = 3 . x = + = π
2 2
The period is 2π . The quarter-period is 2π
or π
. x = π +
π =
3π 2 2
4 2 3π πThe cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
x = π + π
= 3π
2 2
x = 3π
+ π
= 2π 2 2
Evaluate the function at each value of x.
x = + = 2π 2 2
Evaluate the function at each value of x.
2
2
Connect the five key points with a smooth curve and
graph one complete cycle of the given function with
the graph of y = sin x .
Connect the five key points with a smooth curve and graph one complete cycle of the given function with
the graph of y = sin x .
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64 64
Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions
x y = sin 2 x coordinates
0 y = sin 2 ⋅ 0 = sin 0 = 0 (0, 0)
π
4 y = sin
2 ⋅ π
= sin π
= 1 2
π ,1
π
2 y = sin
2 ⋅ π
= sin π = 0
π , 0
2
3π
4 y = sin
2 ⋅
3π
= sin 3π
= −1 2
3π , −1
π y = sin(2 ⋅π ) = sin 2π = 0
(π , 0)
x y = sin 4 x coordinates
0 y = sin(4 ⋅ 0) = sin 0 = 0 (0, 0)
π
8
π π y = sin 4 ⋅ = sin = 1
8 2
π
8
π
4 y = sin
4 ⋅ π
= sin π = 0
4
π , 0
4
3π
8 y = sin
4 ⋅
3π
= sin 3π
= −1 2
3π , −1
8
π
2
y = sin 2π = 0 π 2
, 0
7. The equation y = sin 2 x is of the form y = A sin Bx 8. The equation y = sin 4 x is of the form y = A sin Bx
with A = 1 and B = 2. The amplitude is with A = 1 and B = 4. Thus, the amplitude is
A = 1 = 1 . The period is 2π
= 2π
= π . The
A = 1 = 1 . The period is 2π
= 2π
= π
. TheB 2 B 4 2
πquarter-period is
π . The cycle begins at x = 0. Add
quarter-period is 2 π 1 π
4 quarter-periods to generate x-values for the key points.
x = 0
x = 0 + π 4
x = π
+ π
= π
4 4 2
x = π
+ π
= 3π
= ⋅ = . The cycle begins at 4 2 4 8
x = 0. Add quarter-periods to generate x-values for
the key points. x = 0
x = 0 + π
= π
8 8
x = π
+ π
= π
8 8 4
2 4 4 x =
π + π
= 3π
x = 3π
+ π
= π 4 8 8
3π π π4 4 x = + =
8 8 2Evaluate the function at each value of x. Evaluate the function at each value of x.
4
4 , 1
2
4
4
8
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
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65 65
Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions Section 2.1 Graphs of Sine and Cosine Functions
x 1 y = 3sin x
2
coordinates
0 1 y = 3sin
⋅ 0
2 = 3sin 0 = 3 ⋅ 0 = 0
(0, 0)
π 1 y = 3sin ⋅π
2
= 3sin π
= 3 ⋅1 = 3 2
(π , 3)
2π 1 y = 3sin ⋅ 2π
2 = 3sin π = 3 ⋅ 0 = 0
(2π , 0)
3π 1 y = 3sin ⋅ 3π
2
= 3sin 3π 2
= 3(−1) = −3
(3π , − 3)
4π 1 y = 3sin ⋅ 4π
2 = 3sin 2π = 3 ⋅ 0 = 0
(4π , 0)
π
9. The equation y = 3sin 1
x is of the form 2
y = A sin Bx
10. The equation y = 2sin 1
x is of the form 4
with A = 3 and B = 1
. The amplitude is 2
A = 3
= 3. y = A sin Bx with A = 2 and B = 1
. Thus, the 4
The period is 2π =
2π = 2π ⋅ 2 = 4π . The quarter-
amplitude is A = 2 = 2 . The period is
B 1 2 2
B =
2π = 2π ⋅ 4 = 8π . The quarter-period is
1
period is 4π
= π . The cycle begins at x = 0. Add 4
quarter-periods to generate x-values for the key points. x = 0 x = 0 + π = π x = π + π = 2π x = 2π + π = 3π x = 3π + π = 4π
Evaluate the function at each value of x.
4
8π = 2π . The cycle begins at x = 0. Add quarter-
4 periods to generate x-values for the key points. x = 0 x = 0 + 2π = 2π x = 2π + 2π = 4π x = 4π + 2π = 6π x = 6π + 2π = 8π Evaluate the function at each value of x.
x y = 2sin 1
x 4
coordinates
0 y = 2sin 1
⋅ 0
4
= 2sin 0 = 2 ⋅ 0 = 0
(0, 0)
2π 1 y = 2sin
4 ⋅ 2π
= 2sin π
= 2 ⋅1 = 2 2
(2π , 2)
4π y = 2sin π = 2 ⋅ 0 = 0 (4π , 0)
6π 3π
y = 2sin = 2(−1) = −2 2
(6π , − 2)
8π y = 2sin 2π = 2 ⋅ 0 = 0 (8π , 0)
Connect the five points with a smooth curve and
graph one complete cycle of the given function.
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
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66 66
Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
x y = 4sin π x coordinates
0 y = 4sin(π ⋅ 0) = 4sin 0 = 4 ⋅ 0 = 0
(0, 0)
1
2 y = 4sin
π ⋅
1
= 4sin π
= 4(1) = 4 2
1 , 4
1 y = 4sin(π ⋅1) = 4sin π = 4 ⋅ 0 = 0
(1, 0)
3
2 y = 4sin
π ⋅
3
= 4sin 3π 2
= 4(−1) = −4
3 , − 4
2 y = 4sin(π ⋅ 2) = 4sin 2π = 4 ⋅ 0 = 0
(2, 0)
x y = 3sin 2π x coordinates
0 y = 3sin(2π ⋅ 0) = 3sin 0 = 3 ⋅ 0 = 0
(0, 0)
1
4 y = 3sin
2π ⋅
1
= 3sin π
= 3 ⋅1 = 3 2
1 , 3
1
2 y = 3sin
2π ⋅
1
= 3sin π = 3 ⋅ 0 = 0
1 , 0
3
4
3 y = 3sin 2π ⋅
4
= 3sin 3π
= 3(−1) = −3 2
3 , − 3 4
1 y = 3sin(2π ⋅1) = 3sin 2π = 3 ⋅ 0 = 0
(1, 0)
11. The equation y = 4sin π x is of the form 12. The equation y = 3sin 2π x is of the form
y = A sin Bx with A = 4 and B = π . The amplitude is y = A sin Bx with A = 3 and B = 2π . The amplitude
A = 4 = 4 . The period is 2π
= 2π
= 2 . The
is A = 3 = 3 . The period is 2π
= 2π
= 1 . TheB π B 2π
quarter-period is 2
= 1
. The cycle begins at x = 0. 4 2
Add quarter-periods to generate x-values for the key points.
x = 0
x = 0 + 1
= 1
2 2
x = 1
+ 1
= 1 2 2
x = 1 + 1
= 3
2 2
x = 3
+ 1
= 2 2 2
Evaluate the function at each value of x.
quarter-period is 1
. The cycle begins at x = 0. Add 4
quarter-periods to generate x-values for the key points. x = 0
x = 0 + 1
= 1
4 4
x = 1
+ 1
= 1
4 4 2
x = 1
+ 1
= 3
2 4 4
x = 3
+ 1
= 1 4 4
Evaluate the function at each value of x.
2
2
4
4
2
2
2
2
Connect the five points with a smooth curve and
graph one complete cycle of the given function. Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
66 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 67
x y = −3sin 2π x coordinates
0 y = −3sin(2π ⋅ 0) = −3sin 0 = −3 ⋅ 0 = 0
(0, 0)
1
4 y = −3sin
2π ⋅
1
= −3sin π 2
= −3 ⋅1 = −3
1 , − 3 4
1
2 y = −3sin
2π ⋅
1
= −3sin π = −3 ⋅ 0 = 0
1 , 0 2
3
4 y = −3sin
2π ⋅
3
= −3sin 3π 2
= −3(−1) = 3
3 , 3 4
1 y = −3sin(2π ⋅1) = −3sin 2π = −3 ⋅ 0 = 0
(1, 0)
x y = −2sin π x coordinates
0 y = −2sin(π ⋅ 0) = −2sin 0 = −2 ⋅ 0 = 0
(0, 0)
1
2
1 y = −2sin π ⋅
2
= −2sin π
= −2 ⋅1 = −2 2
1 , − 2 2
1 y = −2sin(π ⋅1) = −2sin π = −2 ⋅ 0 = 0
(1, 0)
3
2 y = −2sin
π ⋅
3
3π
= −2sin = −2(−1) = 2 2
3 , 2
2 y = −2sin(π ⋅ 2) = −2sin 2π = −2 ⋅ 0 = 0
(2, 0)
13. The equation y = −3sin 2π x is of the form 14. The equation y = −2sin π x is of the form
y = A sin Bx with A = –3 and B = 2π . The amplitude y = A sin Bx with A = –2 and B = π . The amplitude
is A = −3 = 3 . The period is 2π
= 2π
= 1 . The
is A = −2 = 2 . The period is 2π
= 2π
= 2 . TheB 2π B π
quarter-period is 1
. The cycle begins at x = 0. Add 4
quarter-periods to generate x-values for the key points.
x = 0
x = 0 + 1
= 1
4 4
x = 1
+ 1
= 1
4 4 2
x = 1
+ 1
= 3
2 4 4
x = 3
+ 1
= 1 4 4
Evaluate the function at each value of x.
quarter-period is 2
= 1
. 4 2
The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x = 0
x = 0 + 1
= 1
2 2
x = 1
+ 1
= 1 2 2
x = 1 + 1
= 3
2 2
x = 3
+ 1
= 2 2 2
Evaluate the function at each value of x.
4
2
4
2
2
Connect the five points with a smooth curve and graph one complete cycle of the given function.
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
66 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 68
x 2 y = − sin x
3
coordinate s
0 2 y = − sin
⋅ 0 3
= − sin 0 = 0
(0, 0)
3π
4 y = − sin
2 ⋅
3π
= − sin π
= −1 2
3π , −1
3π
2 y = − sin
2 ⋅
3π
= − sin π = 0
3π , 0
9π
4 y = − sin
2 ⋅
9π 3 4
= − sin 3π
= −(−1) = 1 2
9π , 1
3π 2 y = − sin ⋅ 3π
3 = − sin 2π = 0
(3π , 0)
x 4
y = − sin x 3
coordinates
0 4
y = − sin ⋅ 0 = − sin 0 = 0 3
(0, 0)
3π
8 y = − sin
4 ⋅
3π = − sin
π = −1
3 8 2
3π , −1
8
3π
4
4 3π y = − sin ⋅ = − sin π = 0
3 4
3π , 0 4
9π
8
4 9π y = − sin ⋅
3 8
= − sin 3π
= −(−1) = 1 2
9π , 1 8
3π
2
4 3π y = − sin ⋅ = − sin 2π = 0
3 2
3π , 0 2
15. The equation y = − sin 2
x is of the form 3
y = A sin Bx
16. The equation y = − sin 4
x is of the form 3
with A = –1 and B = 2
. 3
y = A sin Bx with A = –1 and B = 4
. 3
The amplitude is A = −1 = 1 . The amplitude is A = −1 = 1 .
The period is 2π =
2π = 2π ⋅
3 = 3π .
The period is 2π =
2π = 2π ⋅
3 =
3π .
B 2 2 3
B 4 4 2 3
The quarter-period is 3π
. The cycle begins at x = 0. Add 3π
3π
1 3π4 The quarter-period is 2 = ⋅ = .
quarter-periods to generate x-values for the key points.
x = 0
x = 0 + 3π
= 3π
4 2 4 8
The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x = 0
4 4
x = 3π
+ 3π
= 3π x = 0 +
3π 8
= 3π 8
4 4 2
x = 3π
+ 3π
= 9π
2 4 4
x = 9π
+ 3π
= 3π 4 4
Evaluate the function at each value of x.
x = 3π 8
x = 3π 4
x = 9π 8
+ 3π 8
+ 3π 8
+ 3π 8
= 3π 4
= 9π 8
= 3π 2
Evaluate the function at each value of x.
3 4
4
3 2
2
4
Connect the five points with a smooth curve and
graph one complete cycle of the given function.
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
68 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 68
x y = sin( x − π ) coordinates
π y = sin(π − π ) = sin 0 = 0
(π , 0)
3π
2 y = sin
3π − π
= sin π
= 1 2
3π , 1
2π y = sin(2π − π ) = sin π = 0
(2π , 0)
5π
2 y = sin
5π − π
= sin 3π
= −1 2
5π , − 1
3π y = sin(3π − π ) = sin 2π = 0
(3π , 0)
x π y = sin x −
2
coordinates
π
2
π π y = sin − = sin 0 = 0
2 2
π 2
, 0
π y = sin π − π
= sin π
= 1
2
2
(π , 1)
3π
2 y = sin
3π − π
= sin π = 0 2 2
3π , 0
2
2π π 3π
y = sin 2π − = sin = −1 2 2
(2π , −1)
5π
2
5π π y = sin − = sin 2π = 0
2 2
5π , 0 2
17. The equation y = sin( x − π ) is of the form
18. The equation y = sin x − π
is of the formy = A sin(Bx − C) with A = 1, B = 1, and C = π . The 2
amplitude is A = 1 = 1 . The period is y = A sin(Bx − C) with A = 1, B = 1, and C =
π . The
2π =
2π = 2π . The phase shift is
C = π
= π . The 2
B 1 B 1 amplitude is A = 1 = 1 . The period is
πquarter-period is
2π = π
. The cycle begins at 2π =
2π = 2π . The phase shift is
C = 2 =
π . The
4 2 x = π . Add quarter-periods to generate x-values for
B 1
2π π B 1 2
the key points.
x = π quarter-period is = . The cycle begins at
4 2
x = π + π
= 3π
2 2
x = 3π
+ π
= 2π 2 2
x = 2π + π
= 5π
2 2
x = 5π
+ π
= 3π
x = π
. Add quarter-periods to generate 2
x-values for the key points.
x = π 2
x = π
+ π
= π 2 2
2 2 Evaluate the function at each value of x.
x = π + π
= 3π
2 2
2
2
x = 3π
+ π
= 2π 2 2
x = 2π + π
= 5π
2 2 Evaluate the function at each value of x.
2
2
Connect the five points with a smooth curve and
graph one complete cycle of the given function.
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
69 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 69
x y = sin(2 x − π ) coordinates
π
2 y = sin
2 ⋅ π
− π
= sin(π − π ) = sin 0 = 0
π , 0
3π
4 y = sin
2 ⋅
3π − π
= sin 3π
− π
= sin π
= 1 2
3π , 1
π y = sin(2 ⋅ π − π ) = sin(2π − π ) = sin π = 0
(π , 0)
5π
4 y = sin
2 ⋅
5π − π
= sin 5π
− π
= sin 3π
= −1 2
5π , − 1
3π
2 y = sin
2 ⋅
3π − π
= sin(3π − π ) = sin 2π = 0
2
x y = sin 2 x −
π
coordinates
π
4 y = sin
2 ⋅ π
− π
4 2 π π
= sin − = sin 0 = 0 2 2
π , 0
4
π
2 y = sin
2 ⋅ π
− π
= sin π − π
= sin π
= 1
2
2
π , 1
2
2
19. The equation y = sin(2 x − π ) is of the form Connect the five points with a smooth curve and
y = A sin(Bx − C) with A = 1, B = 2, and C = π . The graph one complete cycle of the given function.
amplitude is A = 1 = 1 . The period is
2π =
2π = π . The phase shift is
C = π
. The
B 2 B 2
quarter-period is π
. The cycle begins at x = π
. Add
4 2quarter-periods to generate x-values for the key points.
x = π 2
x = π
+ π
= 3π
2 4 4
x = 3π
+ π
= π 4 4
20. The equation
y = sin 2 x −
π
is of the form
π
x = π + π
= 5π
y = A sin(Bx − C) with A = 1, B = 2, and C = . The 2
4 4
x = 5π
+ π
= 3π amplitude is A = 1 = 1 .
4 4 2 The period is 2π
= 2π
= π .B 2
Evaluate the function at each value of x. π C π 1 π
= = ⋅ = .The phase shift is
B
2 2 2 4 π
2
2
The quarter-period is . 4
The cycle begins at x = π
. Add quarter-periods to 4
generate x-values for the key points.
π 4
4 x =
4 π π π
2
x = + = 4 4 2
x = π
+ π
= 3π
2 4 4 3π π
x = + = π 4 4
x = π + π
= 5π
4 4
4
4
Evaluate the function at each value of x.
2
2
2
3π , 0
2 2
2
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
70 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 70
3π
4 y = sin
2 ⋅
3π − π
= sin 3π
− π
2 2
= sin π = 0
3π , 0
π y = sin 2 ⋅ π −
π
= sin 2π −
π
= sin 3π
= −1 2
(π , − 1)
5π
4 y = sin
2 ⋅
5π − π
= sin 5π
− π
2 2
= sin 2π = 0
5π , 0
x y = 3sin(2x − π ) coordinates
π
2
π y = 3sin 2 ⋅ − π
2 = 3sin(π − π ) = 3sin 0 = 3 ⋅ 0 = 0
π 2
, 0
3π
4 y = 3sin
2 ⋅
3π − π
= 3sin 3π
− π 2
= 3sin π
= 3 ⋅1 = 3 2
3π , 3
4
π y = 3sin(2 ⋅π − π ) = 3sin(2π − π ) = 3sin π = 3 ⋅ 0 = 0
(π , 0)
5π
4
5π y = 3sin 2 ⋅ − π
4
= 3sin 5π
− π
= 3sin 3π 2
= 3(−1) = −3
5π
4 , − 3
3π
2 y = 3sin
2 ⋅
3π − π
= 3sin(3π − π ) = 3sin 2π = 3 ⋅ 0 = 0
3π , 0
2
4 2
4
2
2
4
4 2
4
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
2
2
2
Connect the five points with a smooth curve and
graph one complete cycle of the given function.
21. The equation y = 3sin(2x − π ) is of the form
y = A sin(Bx − C ) with A = 3, B = 2, and C = π . The
amplitude is A = 3 = 3 . The period is
2π =
2π = π . The phase shift is
C = π
. The quarter-
B 2 B 2
period is π
. The cycle begins at x = π
. Add quarter-
4 2periods to generate x-values for the key points.
x = π
22. The equation
y = 3sin 2 x −
π
is of the form
2
2
x = π
+ π
= 3π
2 4 4
x = 3π
+ π
= π
y = A sin(Bx − C ) with A = 3, B = 2, and C = π
. 2
4 4
x = π + π
= 5π
The amplitude is
2π A = 3
2π = 3 .
4 4
x = 5π
+ π
= 3π
4 4 2
The period is = = π . B 2
C π
π 1 π
= = ⋅ = .Evaluate the function at each value of x.
The phase shift is B
2 2 2 4
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
71 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 71
x y = 3sin 2 x −
π
coordinates
π
4 y = 3sin
2 ⋅ π
− π
= sin π
− π
= 3sin 0 = 3 ⋅ 0 = 0
π , 0
π
2 y = 3sin
2 ⋅ π
− π
= 3sin π − π
= 3sin π
= 3 ⋅1 = 3 2
π , 3
3π
4 y = 3sin
2 ⋅
3π − π
= 3sin 3π
− π
2 2
= 3sin π = 3 ⋅ 0 = 0
4
π y = 3sin 2 ⋅ π −
π
2
= 3sin 2π −
π
2
= 3sin 3π
= 3 ⋅ (−1) = −3 2
(π , − 3)
5π
4 y = 3sin
2 ⋅
5π − π
= 3sin 5π
− π
2 2
= 3sin 2π = 3 ⋅ 0 = 0
5π , 0
−
x
= −
π
The quarter-period is π
. 4
The cycle begins at x = π
. Add quarter-periods to 4
generate x-values for the key points.
x = π 4
x = π
+ π
= π
4 4 2
x = π
+ π
= 3π
2 4 4
x = 3π
+ π
= π
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
4 4 1 π 1 π x = π +
π =
5π 23. y = sin x + = sin x − − 2 2 2 2
4 4
Evaluate the function at each value of x.
The equation
y = 1
sin x −
− π
is of the form2
2
2
1 πy = A sin(Bx − C ) with A = , B = 1, and C = − .
4 2
4
The amplitude is
2 2
A = 1
= 1
. The period is 2 2
2 2
2π 2π π
C 2= = 2π . The phase shift is B 1
2π π
= = − . B 1 2
2 2
2
The quarter-period is
π
= . The cycle begins at 4 2
2 x = − . Add quarter-periods to generate x-values
2 for the key points.
π = −
4 2
3π , 0
2
x π
+ π
= 0 2 2
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
x = π + π
= 3π
2 2 Evaluate the function at each value of x.
4 2
4
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
72 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 72
x y = 1
sin x + π
2
2
coordinates
− π 2
y = 1
sin − π
+ π
2
2 2
= 1
sin 0 = 1
⋅ 0 = 0 2 2
2
0 y = 1
sin 0 + π
2
2
= 1
sin π
= 1
⋅1 = 1
2 2 2 2
0,
1
π
2 y =
1 sin
π + π
2 2 2
= 1
sin π = 1
⋅ 0 = 0 2 2
2
π y = 1
sin π + π
2
2
= 1
sin 3π
2 2
= 1
⋅ (−1) = − 1
2 2
π , −
1
3π
2 y =
1 sin
3π + π
2
2 2
= 1
sin 2π 2
= 1
⋅ 0 = 0 2
3π , 0
x 1
y = sin( x + π ) 2
coordinates
−π 1
y = sin(−π + π ) 2 1 1
= sin 0 = ⋅ 0 = 0 2 2
(−π , 0)
π −
2
1 π y = sin − + π
2 2
= 1
sin π
= 1
⋅1 = 1
2 2 2 2
π 1 −
2 ,
2
0 y = 1
sin(0 + π ) 2 1 1
= sin π = ⋅ 0 = 0 2 2
(0, 0)
π
2 y =
1 sin
π + π
2 2
= 1
sin 3π
= 1
⋅ (−1) = − 1
2 2 2 2
π , −
1 2 2
π y = 1
sin(π + π ) 2 1 1
= sin 2π = ⋅ 0 = 0 2 2
(π , 0)
2π
= 2π
= 2π . The phase shift is C
= −π
= −π . B 1 B 1
− π
, 0
The quarter-period is 2π
= π
. The cycle begins at 4 2
x = −π . Add quarter-periods to generate x-values for the key points. x = −π
x = −π + π
= − π
2 2 π π
2
x = − + = 0 2 2
x = 0 + π
= π
2 2
x = π
+ π
= π π
, 0 2 2 Evaluate the function at each value of x.
2
2
Connect the five points with a smooth curve and
graph one complete cycle of the given function.
24.
y = 1
sin( x + π ) = 1
sin( x − (−π )) 2 2
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
The equation y = 1
sin( x − (−π )) 2
is of the form
y = A sin(Bx − C) with A = 1
, B = 1, and C = −π . 2
The amplitude is A = 1 =
1 . The period is
2 2
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
73 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 73
π
2 y = −2sin
2 ⋅ π
+ π
2 2 π
= −2sin π +
3π 2
= −2sin 2
= −2(−1) = 2
π , 2 2
3π
4 y = −2sin
2 ⋅
3π + π
= −2sin 3π
+ π
2 2 = −2sin 2π = −2 ⋅ 0 = 0
3π , 0
4
x y = −2sin
2 x +
π
coordinates
− π 4
y = −2sin 2⋅
− π
+ π
= −2sin − π
+ π
= −2sin 0 = −2 ⋅ 0 = 0
− π
, 0
0 y = −2sin
2 ⋅ 0 +
π
= −2sin 0 + π
= −2sin π 2
= −2 ⋅1 = −2
(0, –2)
π
4 y = −2sin
2 ⋅ π
+ π
= −2sin π
+ π
2 2
= −2sin π = −2 ⋅ 0 = 0
4
= −
= −
= −
C
25.
y = −2sin 2 x +
π = −2 sin
2x −
− π
2
2
The equation
y = −2sin 2 x −
− π
is of the form 2
y = A sin(Bx − C) with A = –2,
B = 2, and C π
. The amplitude is
A = −2
2
= 2 . The period is 2π
= 2π
= π . The
4 2
B 2 π
phase shift is C
= −
2 = − π
⋅ 1
= − π
. The quarter- B 2 2 2 4
period is π
. The cycle begins at x π
. Add Connect the five points with a smooth curve and4 4 graph one complete cycle of the given function.
quarter-periods to generate x-values for the key
points.
π x = −
4 π π
x = − + = 0 4 4
x = 0 + π
= π
4 4
x = π
+ π
= π
4 4 2
x = π
+ π
= 3π
2 4 4
26.
y = −3sin 2x +
π = −3sin
2 x −
− π
Evaluate the function at each value of x.
2
2
π The equation y = −3sin 2 x − −
is of the form
2 2
π
4
2
4
y = A sin(Bx − C) with A = –3, B = 2, and C = − . 2
The amplitude is A = −3 = 3 . The period is
2 2
2π =
2π = π . The phase shift is
B 2
π−
2 π 1 π π
2 = = − ⋅ = − . The quarter-period is . B 2 2 2 4 4
π 2 The cycle begins at x = − . Add quarter-periods to
4 generate x-values for the key points.
π x = −
4 2 π
, 0
4
x π
+ π
= 0 4 4
x = 0 + π
= π
4 4
x = π
+ π
= π
4 4 2
x = π
+ π
= 3π
2 4 4
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
74 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 74
x y = −3sin 2 x +
π
coordinates
− π 4
y = −3sin 2 ⋅
− π
+ π
= −3sin − π
+ π
2 2
= −3sin 0 = −3 ⋅ 0 = 0
− π
, 0
0 y = −3sin 2 ⋅ 0 +
π
2
= −3sin 0 + π
= −3sin π
= −3 ⋅1 = −3 2
(0, –3)
π
4 y = −3sin
2 ⋅ π
+ π
= −3sin π
+ π
2 2
= −3sin π = −3 ⋅ 0 = 0
π , 0
π
2 y = −3sin
2 ⋅ π
+ π
= −3sin π + π
= −3sin 3π
= −3 ⋅ (−1) = 3 2
π , 3
3π
4 y = −3sin
2 ⋅
3π + π
= −3sin 3π
+ π
= −3sin 2π = −3 ⋅ 0 = 0
3π , 0
x y = 3sin(π x + 2) coordinates
− 2 π
y = 3sin π
−
2 + 2
= 3sin(−2 + 2) = 3sin 0 = 3 ⋅ 0 = 0
2 − , 0 π
π − 4
2π
π − 4 y = 3sin π + 2
2π
= 3sin π − 4
+ 2 2
π = 3sin − 2 + 2
2
= 3sin π 2
= 3 ⋅1 = 3
π − 4 2π
π − 2
π
π − 2 y = 3sin π
π + 2
= 3sin(π − 2 + 2) = 3sin π = 3 ⋅ 0 = 0
π − 2 , 0
3π − 4
2π
3π − 4 y = 3sin π + 2
2π
= 3sin 3π − 4
+ 2 2
= 3sin 3π
− 2 + 2 2
= 3sin 3π 2
= 3(−1) = −3
5π , − 3 4
2π − 2
π y = 3sin
π
2π − 2 + 2
= 3sin(2π − 2 + 2) = 3sin 2π = 3 ⋅ 0 = 0
2π − 2 , 0
Evaluate the function at each value of x. 2π =
2π = 2 . The phase shift is C
= −2
= − 2
. The
2
B π B π π 2 1
quarter-period is = . The cycle begins at 4 2
4
2
4
x = − 2
. Add quarter-periods to generate x-values π
for the key points.
= − 2
x π
= − 2
+ 1
= π − 4
x π 2 2π
= π − 4
+ 1
= π − 2
x 2π 2 π
2 x = π − 2
+ 1
= 3π − 4
π 2 2π
x = 3π − 4
+ 1
= 2π − 2
2π 2 π
4 2
4
Evaluate the function at each value of x.
2 2
2
2
π
, 3
4 2
2 2
4
27.
Connect the five key points with a smooth curve and graph one complete cycle of the given function.
y = 3sin(π x + 2)
π
The equation y = 3sin(π x − (−2)) is of the form
y = A sin(Bx − C) with A = 3, B = π , and C = –2.
The amplitude is A = 3 = 3 . The period is
π π
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
75 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 75
π − 4
2π
π − 4 y = 3sin 2π + 4
2π = 3sin(π − 4 + 4) = 3sin π = 3 ⋅ 0 = 0
π − 4
2π , 0
3π − 8
4π
3π − 8 y = 3sin 2π + 4
4π
= 3sin 3π − 8
+ 4 2
= 3sin 3π
− 4 + 4 2
= 3sin 3π
= 3(−1) = −3 2
3π − 8 , − 3
4π
π − 2
π
π − 2 y = 3sin 2π
π + 4
= 3sin(2π 4 + 4)
= π = 3 ⋅ 0 = 0 3sin 2
π − 2 ,0
x y = 3sin(2π x + 4) coordinates
2 − π
2 y = 3sin 2π − + 4
π = 3sin(−4 + 4) = 3sin 0 = 3 ⋅ 0 = 0
2 − π
, 0
π − 8
4π
π − 8 y = 3sin 2π + 4
4π
= 3sin π − 8
+ 4
= 3sin π
− 4 + 4
= 3sin π
= 3 ⋅1 = 3 2
π − 8 , 3
4π
= −
Connect the five points with a smooth curve and
graph one complete cycle of the given function.
28. y = 3sin(2π x + 4) = 3sin(2π x − (−4))
The equation y = 3sin(2π x − (−4)) is of the form
y = A sin(Bx − C) with A = 3, B = 2π , and
C = –4. The amplitude is
A = 3
= 3 . The period − π
is 2π
= 2π
= 1 . The phase shift is C
= −4
= − 2
.B 2π B 2π π Connect the five key points with a smooth curve and
The quarter-period is 1
. The cycle begins at 4
x = − 2
. Add quarter-periods to generate x-values π
for the key points.
x = − 2 π
x = − 2
+ 1
= π − 8
graph one complete cycle of the given function.
π 4 4π
x = π − 8
+ 1
= π − 4
4π 4 2π 29. y = −2 sin(2π x + 4π ) = −2 sin(2π x − (−4π ))x = π − 4
+ 1
= 3π − 8
The equation
y = −2sin(2π x − (−4π )) is of the form2π 4 4π
x = 3π − 8
+ 1
= π − 2 y = A sin(Bx − C) with A = –2, B = 2π , and
4π 4 π C = −4π . The amplitude is A = −2 = 2 . TheEvaluate the function at each value of x.
period is 2π
= 2π
= 1 . The phase shift is B 2π
C −4π 1 = = −2 . The quarter-period is . The cycle
B 2π 4
2
begins at x = −2 . Add quarter-periods to generate x-
values for the key points.
x = −2
x = −2 + 1
= − 7
4 4
x 7
+ 1
= − 3
4 4 2 3 1 5
2
x = − + = −
2 4 4 5 1
= − + = −x 1
4 4
Evaluate the function at each value of x.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
76 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 76
x y = −2sin(2π x + 4π ) coordinates
–2 y = −2sin(2π (−2) + 4π ) = −2sin(−4π + 4π ) = −2sin 0 = −2 ⋅ 0 = 0
(–2, 0)
− 7 4
y = −2sin 2π
−
7 + 4π
= −2sin −
7π + 4π
2
= −2sin π
= −2 ⋅1 = −2 2
−
7 , − 2
− 3 2
y = −2sin 2π
−
3 + 4π
= −2sin(−3π + 4π ) = −2sin π = −2 ⋅ 0 = 0
−
3 , 0
− 5 4
y = −2sin 2π
−
5 + 4π
= −2sin −
5π + 4π
= −2sin 3π 2
= −2(−1) = 2
5 , 2
4
–1 y = −2sin(2π (−1) + 4π ) = −2sin(−2π + 4π ) = −2sin 2π = −2 ⋅ 0 = 0
(–1, 0)
x y = −3sin(2π x + 4π ) coordinates
–2 y = −3sin(2π (−2) + 4π ) = −3sin(−4π + 4π ) = −3sin 0 = −3 ⋅ 0 = 0
(–2, 0)
7 −
4
7 y = −3sin 2π − + 4π
4 π
= −3sin − 7
+ 4π 2
= −3sin π
= −3 ⋅1 = −3 2
7 −
4 , − 3
3 −
2
3 y = −3sin 2π − + 4π
2 = −3sin(−3π + 4π ) = −3sin π = −3 ⋅ 0 = 0
3 − , 0 2
5 −
4
5 y = −3sin 2π − 4
+ 4π
= −3sin −
5π + 4π
= −3sin 3π
= −3(−1) = 3 2
−
5 , 3
4
–1 y = −3sin(2π (−1) + 4π ) = −3sin(−2π + 4π ) = −3sin 2π = −3 ⋅ 0 = 0
(–1, 0)
= −
= −
x = −2
x = −2 + 1
= − 7
4 4 7
x = − + 1
= − 3
4 4 2
4
4
x 3
+ 1
= − 5
2 4 4
x 5
+ 1
= −1 4 4
Evaluate the function at each value of x.
2
2
4
2
−
Connect the five points with a smooth curve and
graph one complete cycle of the given function.
2
30.
y = −3sin(2π x + 4π ) = −3sin(2π x − (−4π ))
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
The equation y = −3sin(2π x − (−4π )) is of the form
y = A sin(Bx − C ) with A = –3, B = 2π , and
C = −4π . The amplitude is A = −3 = 3 . The
period is 2π
= 2π
= 1 . The phase shift is B 2π
C =
−4π = −2 . The quarter-period is
1 . The cycle
B 2π 4
begins at x = −2 . Add quarter-periods to generate x-
values for the key points.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
77 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 77
x y = 2 cos x coordinates
0 y = 2 cos 0 = 2 ⋅1 = 2 (0, 2)
π
2 y = 2 cos
π = 2 ⋅ 0 = 0
2
2
π y = 2 cos π = 2 ⋅ (−1) = −2
(π , − 2)
3π
2 y = 2 cos
3π 2
= 2 ⋅ 0 = 0
2
2π y = 2 cos 2π = 2 ⋅1 = 2
(2π , 2)
x y = 3cos x coordinates
0 y = 3cos 0 = 3 ⋅1 = 3 (0, 3)
π
2
π y = 3cos = 3 ⋅ 0 = 0
2
π
2
π y = 3cos π = 3 ⋅ (−1) = −3 (π , − 3)
3π
2 y = 3cos
3π = 3 ⋅ 0 = 0
2
3π , 0
2
2π y = 3cos 2π = 3 ⋅1 = 3 (2π , 3)
31. The equation y = 2 cos x is of the form y = A cos x 32. The equation y = 3cos x is of the form y = A cos x
with A = 2. Thus, the amplitude is A = 2 = 2 . with A = 3. Thus, the amplitude is A = 3 = 3 .
The period is 2π . The quarter-period is 2π or
π . The period is 2π . The quarter-period is
2π or π
.4 2
The cycle begins at x = 0 . Add quarter-periods to
generate x-values for the key points.
x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
x = π + π
= 3π
4 2 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
x = π + π
= 3π
2 2 3π
2 2
x = 3π
+ π
= 2π 2 2
Evaluate the function at each value of x.
x = + π
= 2π 2 2
Evaluate the function at each value of x.
π , 0 , 0
3π , 0
Connect the five key points with a smooth curve
and graph one complete cycle of the given function
Connect the five points with a smooth curve and
graph one complete cycle of the given function with
with the graph of y = cos x .
the graph of y = 2 cos x .
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
78 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 78
x y = −3cos x coordinates
0 y = −3cos 0 = −3 ⋅1 = −3 (0, –3)
π
2
π y = −3cos
2 = −3 ⋅ 0 = 0
π 2
, 0
π y = −3cos π = −3 ⋅ (−1) = 3 (π , 3)
3π
2
3π y = −3cos = −3 ⋅ 0 = 0
2
3π , 0 2
2π y = −3cos 2π = −3 ⋅1 = −3 (2π , − 3)
x y = −2 cos x coordinates
0 y = −2 cos 0 = −2 ⋅1 = −2
(0, –2)
π
2 y = −2 cos
π 2
= −2 ⋅ 0 = 0
π , 0
2
π y = −2 cos π = −2 ⋅ (−1) = 2
(π , 2)
3π
2 y = −2 cos
3π 2
= −2 ⋅ 0 = 0
3π , 0
2
2π y = −2 cos 2π = −2 ⋅1 = −2
(2π , − 2)
33. The equation y = −2 cos x is of the form y = A cos x x = 0
with A = –2. Thus, the amplitude is x = 0 + π
= π
A = −2 = 2 . The period is 2π . The quarter-
2 2
π
period is 2π
or π
. The cycle begins at x = 0 . Add
x = + π
= π 2 2
4 2 quarter-periods to generate x-values for the key points.
x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
x = π + π
= 3π
2 2
x = 3π
+ π
= 2π 2 2
Evaluate the function at each value of x.
x = π + π
= 3π
2 2
x = 3π
+ π
= 2π 2 2
Evaluate the function at each value of x.
Connect the five key points with a smooth curve and
graph one complete cycle of the given function with
the graph of y = cos x .
Connect the five points with a smooth curve and 35. The equation y = cos 2 x is of the form y = A cos Bx
graph one complete cycle of the given function with with A = 1 and B = 2. Thus, the amplitude isthe graph of y = cos x .
A = 1
= 1 . The period is 2π
= 2π
= π . The B 2
34. The equation
y = −3cos x is of the form
y = A cos x
quarter-period is π
. The cycle begins at x = 0 . Add 4
quarter-periods to generate x-values for the key points.
x = 0
x = 0 + π
= π
4 4 π π π
with A = –3. Thus, the amplitude is A = −3 = 3 . x = + = 4 4 2
The period is 2π . The quarter-period is 2π or
π . x =
π + π
= 3π
4 2 The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points.
2 4 4
x = 3π
+ π
= π 4 4
Evaluate the function at each value of x.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
79 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 79
x y = cos 2 x coordinates
0 y = cos(2 ⋅ 0) = cos 0 = 1
(0, 1)
π
4 y = cos
2 ⋅ π
= cos π
= 0 2
π , 0
π
2 y = cos
2 ⋅ π
= cos π = −1
π , −1
3π
4 y = cos
2 ⋅
3π
= cos 3π
= 0 2
3π , 0
π y = cos(2 ⋅ π ) = cos 2π = 1
(π , 1)
x y = cos 4 x coordinates
0 y = cos(4 ⋅ 0) = cos 0 = 1 (0, 1)
π
8
π π y = cos 4 ⋅
8 = cos
2 = 0
π 8
, 0
π
4 y = cos
4 ⋅ π
= cos π = −1
π , −1
3π
8
3π y = cos 4 ⋅
8
= cos 3π
= 0 2
3π 8
, 0
π
2
π y = cos 4 ⋅
2 = cos 2π = 1
π 2
, 1
4
4
2
2
4
4
4
4
Connect the five points with a smooth curve and graph one complete cycle of the given function.
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
37. The equation y = 4 cos 2π x is of the form y = A cos Bx
with A = 4 and B = 2π . Thus, the amplitude is
A = 4 = 4 . The period is 2π
= 2π
= 1 . The quarter-B 2π
136. The equation y = cos 4 x is of the form y = A cos Bx period is . The cycle begins at x = 0 . Add quarter-
4with A = 1 and B = 4. Thus, the amplitude is periods to generate x-values for the key points.
A = 1
= 1 . The period is 2π
= 2π
= π
. The x = 0
B 4 2 π
x = 0 + 1
= 1
4 4
quarter-period is 2 = π
⋅ 1
= π
. The cycle begins at 4 2 4 8
x = 0. Add quarter-periods to generate x-values for
the key points. x = 0
x = 0 + π
= π
8 8
x = π
+ π
= π
8 8 4
x = π
+ π
= 3π
4 8 8
x = 3π
+ π
= π
8 8 2 Evaluate the function at each value of x.
x = 1
+ 1
= 1
4 4 2
x = 1
+ 1
= 3
2 4 4
x = 3
+ 1
= 1 4 4
Evaluate the function at each value of x.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
80 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 80
x y = 4 cos 2π x coordinates
0 y = 4 cos(2π ⋅ 0) = 4 cos 0 = 4 ⋅1 = 4
(0, 4)
1
4 y = 4 cos
2π ⋅
1
= 4 cos π 2
= 4 ⋅ 0 = 0
1 , 0
1
2 y = 4 cos
2π ⋅
1
= 4 cos π = 4 ⋅ (−1) = −4
1 , − 4
3
4 y = 4 cos
2π ⋅
3
= 4 cos 3π 2
= 4 ⋅ 0 = 0
3 , 0
1 y = 4 cos(2π ⋅1) = 4 cos 2π = 4 ⋅1 = 4
(1, 4)
x y = 5cos 2π x coordinates
0 y = 5 cos(2π ⋅ 0) = 5cos 0 = 5 ⋅1 = 5
(0, 5)
1
4 y = 5cos
2π ⋅
1 4
= 5cos π
= 5 ⋅ 0 = 0 2
1 , 0
4
1
2 y = 5cos
2π ⋅
1 2
= 5cos π = 5 ⋅ (−1) = −5
1 , − 5
2
3
4 y = 5cos
2π ⋅
3π 4
= 5cos 3π
= 5 ⋅ 0 = 0 2
3 , 0
4
1 y = 5cos(2π ⋅1) = 5cos 2π = 5 ⋅1 = 5
(1, 5)
4
4
2
2
4
4
Connect the five points with a smooth curve and
graph one complete cycle of the given function.
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
38. The equation
y = 5cos 2π x
is of the form
39. The equation y = −4 cos 1
x is of the form 2
1y = A cos Bx with A = 5 and B = 2π . Thus, the y = A cos Bx with A = –4 and B = . Thus, the
2amplitude is A = 5 = 5 . The period is amplitude is A = −4 = 4 . The period is
2π =
2π = 1 . The quarter-period is
1 . The cycle 2π
= 2π
= 2π ⋅ 2 = 4π . The quarter-period isB 2π 4 B 1
begins at x = 0. Add quarter-periods to generate x-
values for the key points. x = 0
x = 0 + 1
= 1
4 4
x = 1
+ 1
= 1
4 4 2
x = 1
+ 1
= 3
2 4 4
x = 3
+ 1
= 1 4 4
Evaluate the function at each value of x.
2
4π = π . The cycle begins at x = 0 . Add quarter-
4 periods to generate x-values for the key points. x = 0 x = 0 + π = π x = π + π = 2π x = 2π + π = 3π x = 3π + π = 4π
Evaluate the function at each value of x.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
81 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 81
x 1 y = −4 cos x
2
coordinates
0 y = −4 cos
1 ⋅ 0
= −4 cos 0 = −4 ⋅1 = −4
(0, –4)
π 1 y = −4 cos ⋅ π
2
= −4 cos π 2
= −4 ⋅ 0 = 0
(π , 0)
2π 1 y = −4 cos ⋅ 2π
2 = −4 cos π = −4 ⋅ (−1) = 4
(2π , 4)
3π 1 y = −4 cos ⋅ 3π
2
= −4 cos 3π 2
= −4 ⋅ 0 = 0
(3π , 0)
4π 1 y = −4 cos ⋅ 4π
2 = −4 cos 2π = −4 ⋅1 = −4
(4π , – 4)
x y = −3cos
1 x
3
coordinates
0 y = −3cos
1 ⋅ 0
3 = −3cos 0 = −3 ⋅1 = −3
(0, –3)
3π
2 y = −3cos
1 ⋅
3π
= −3cos π
= −3 ⋅ 0 = 0 2
3π , 0
3π y = −3cos
1 ⋅ 3π
3 = −3cos π = −3 ⋅ (−1) = 3
(3π , 3)
9π
2 y = −3cos
1 ⋅
9π
3π = −3cos
2 = −3 ⋅ 0 = 0
9π , 0
6π y = −3cos
1 ⋅ 6π 3
= −3cos 2π = −3 ⋅1 = −3
(6π , − 3)
= − 3
x = 0
3π 3π
2
x = 0 + = 2 2
x = 3π
+ 3π
= 3π 2 2
x = 3π + 3π
= 9π
2 2
x = 9π
+ 3π
= 6π 2 2
Evaluate the function at each value of x.
3 2
2
Connect the five points with a smooth curve and
graph one complete cycle of the given function.
3 2
2
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
40. The equation
y = A cos Bx
y = −3cos 1
x is of the form 3
with A = –3 and B = 1
. Thus, the 3
amplitude is A = −3 = 3 . The period is
2π =
2π = 2π ⋅ 3 = 6π . The quarter-period is
41. The equation
y
1 cos π
x is of the formB 1
2 3
6π =
3π . The cycle begins at x = 0. Add quarter-
4 2
y = A cos Bx
with 1
A = − 2
and B = π
. Thus, the 3
periods to generate x-values for the key points.
amplitude is A = − 1
= 1
. The period is 2 2
2π =
2π = 2π ⋅
3 = 6 . The quarter-period is
6 =
3 .
B π π 4 2 3
The cycle begins at x = 0 . Add quarter-periods to
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82 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 82
x y 1
cos π
x 2 3
coordinates
0 y 1
cos π
⋅ 0 2 3 1
cos 0 2 1
⋅1 = − 1
2 2
0, −
1 2
3
2 y
1 cos
π 3 2 3 2 1
cos π = −
2 2 1
⋅ 0 = 0 = −
2
3 , 0 2
3 y 1
cos π
⋅ 3 2 3 1
cos π = −
2 1
⋅ (−1) = 1
2 2
3,
1 2
9
2 y
1 cos
π 9 2 3 2 1
cos 3π
2 2 1
⋅ 0 = 0 2
9 , 0 2
6 y 1
cos π
⋅ 6 2 3 1
cos 2π 2 1
⋅1 = − 1
2 2
6, −
1
2
x y
1 cos π
x
= − 2 4
coordinates
0 1 π y
2 cos
4 ⋅ 0 = −
1
cos 0 1
⋅1 = − 1 = − = −
2 2 2
1 0, −
2
2 y
1 cos
π ⋅ 2
= − 2 4
1 cos π
= − 1
⋅ 0 = 0 2 2 2
(2, 0)
4 1 π y
2 cos
4 ⋅ 4 = −
1 1 1 = − cos π = − ⋅ (−1) =
2 2 2
1 4,
2
6 y
1 cos
π ⋅ 6
2 4 1
cos 3π
= − 1
⋅ 0 = 0
= − 2 2 2
(6, 0)
8 1 π y = − cos ⋅ 8
2 4 1
cos 2π = − 1
⋅1 = − 1 = −
2 2 2
1 8, − 2
= −
generate x-values for the key points.
x = 0
x = 0 + 3
= 3
2 2
x = 3
+ 3
= 3 2 2
x = 3 + 3
= 9
2 2
x = 9
+ 3
= 6 2 2
Evaluate the function at each value of x.
Connect the five points with a smooth curve and
graph one complete cycle of the given function.
= − 42. The equation y
1 cos π
x 2 4
1
is of the form
π
= −
y = A cos Bx with A = − and B = . Thus, the 2 4
1 1
= − amplitude is
A = − = . The period is
2 2
= − 2π
= 2π
B π 4
= 2π ⋅ 4
= 8 . The quarter-period is π
8 = 2 . 4
= − ⋅ The cycle begins at x = 0. Add quarter-periods to
generate x-values for the key points. x = 0 x = 0 + 2 = 2 x = 2 + 2 = 4 x = 4 + 2 = 6 x = 6 + 2 = 8
Evaluate the function at each value of x.
= −
= −
= − ⋅
= −
= − = −
= −
= −
= −
= −
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83 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 83
x coordinates
− π 2
− π
, 1
0 (0, 0)
π
2
π , −1 2
π (π , 0)
3π
2
3π , 1
x coordinates
π
2
π , 1
π (π , 0)
3π
2
3π , −1
2
2π (2π , 0)
5π
2
5π , 1 2
x
= −
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
Connect the five points with a smooth curve and
graph one complete cycle of the given function
π
43. The equation
y = cos x − π
is of the form 44. The equation y = cos x +
2 is of the form
2
y = A cos ( Bx − C ) with A = 1, and B = 1, and y = A cos ( Bx − C ) with A = 1, and B = 1, and
π
C = π
. Thus, the amplitude is A = 1 = 1 . The C = − . Thus, the amplitude is 2
A = 1 = 1 . The
2
period is 2π
= 2π
= 2π . The phase shift is
period is
2π =
2π B 1
= 2π . The phase shift is
B 1 π
π −C 2 π 2π π
C = 2 =
π . The quarter-period is
2π = π
. The = = − . The quarter-period is B 1 2
= . The 4 2
B 1 2 4 2 π
cycle begins at x = π
. Add quarter-periods to 2
generate x-values for the key points.
x = π 2
x = π
+ π
= π 2 2
x = π + π
= 3π
2 2
x = 3π
+ π
= 2π
cycle begins at x = − . Add quarter-periods to 2
generate x-values for the key points.
π = −
2
x π
+ π
= 0 2 2
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
2 2
x = 2π + π
= 5π x = π +
π =
3π 2 2
2 2 Evaluate the function at each value of x.
Evaluate the function at each value of x.
2
2
2
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84 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 84
x coordinates
π
2
π 2
, 4
3π
4
3π , 0
4
π (π , − 4)
5
4
5π , 0
4
3π
2
3π , 4
2
x coordinates
π
2
π , 3
3π
4
3π , 0
4
π (π , − 3)
5π
4
5π
4 , 0
3π
2
3π , 3 2
Connect the five points with a smooth curve and
graph one complete cycle of the given function
Connect the five points with a smooth curve and
graph one complete cycle of the given function
45. The equation
y = 3cos(2 x − π ) is of the form 46. The equation y = 4 cos(2 x − π ) is of the form
y = A cos ( Bx − C ) with A = 3, and B = 2, and y = A cos(Bx − C ) with A = 4, and B = 2, and C = π .
C = π . Thus, the amplitude is
A = 3
= 3 . The Thus, the amplitude is A = 4 = 4 . The period is
2π =
2π = π . The phase shift is
C = π
. The
period is 2π
= 2π
= π . The phase shift is C
= π
. B 2 B 2 B 2 B 2 π π
The quarter-period is π
. The cycle begins at x = π
. quarter-period is . The cycle begins at x = . Add
4 24 2 quarter-periods to generate x-values for the key points.
Add quarter-periods to generate x-values for the key points. x =
π 2
x = π 2
x = π
+ π
= 3π
2 4 4
x = 3π
+ π
= π 4 4
x = π + π
= 5π
4 4
x = 5π
+ π
= 3π
4 4 2 Evaluate the function at each value of x.
x = π
+ π
= 3π
2 4 4
x = 3π
+ π
= π 4 4
x = π + π
= 5π
4 4
x = 5π
+ π
= 3π
4 4 2
Evaluate the function at each value of x.
2
π
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85 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 85
x coordinates
− π 6
− π
, 1
6 2
0 (0, 0)
π
6
π , −
1 6 2
π
3
π , 0
3
π
2
π ,
1 2 2
= −
= −
2
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
47.
y = 1
cos 3x +
π =
1 cos
3x −
− π
2
2
2
2
Connect the five points with a smooth curve and
The equation
y = 1
cos 3x −
− π
is of the form graph one complete cycle of the given function
2
2
y = A cos(Bx − C ) with
C π
A = 1
, and B = 3, and 2
1 1= − . Thus, the amplitude is
2 A = = . The
2 2
period is 2π
= 2π
. The phase shift is
B 3 π
C =
− 2 = −
π ⋅ 1
= − π
. The quarter-period is 1 1B 3 2 3 6
2π
48. y = cos(2 x + π ) = cos(2 x − (−π )) 2 2
3 = 2π
⋅ 1
= π
. The cycle begins at x π
. Add
The equation y = 1
cos(2x − (−π ))
is of the form4 3 4 6 6 2
quarter-periods to generate x-values for the key points.
y = A cos(Bx − C ) with
A = 1
, and B = 2, andπ
x = − 6 π π
x = − + = 0 6 6
C = −π . Thus, the amplitude is A = 1
= 1
. The 2 2
x = 0 + π
= π period is
2π =
2π = π . The phase shift is
6 6 B 2
x = π
+ π
= π C
= −π
= − π
. The quarter-period is π
. The cycle6 6 3
x = π
+ π
= π B 2 2 4
3 6 2
Evaluate the function at each value of x.
begins at x π
. Add quarter-periods to generate x- 2
values for the key points.
π x = −
2 π π π
x = − + = − 2 4 4 π π
x = − + = 0 4 4
x = 0 + π
= π
4 4
x = π
+ π
= π
4 4 2 Evaluate the function at each value of x.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
86 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 86
x
coordinates
− π 2
− π
, 1
− π 4
− π
, 0
0 0, −
1
4
4
π
2
π , 1
x
coordinates
π
4
π , − 3 4
2
2
3π
4
3π
4 , 3
π
(π , 0)
5π
4
5π , − 3 4
Evaluate the function at each value of x.
2 2
4
2
π π , 0
π π , 0
2 2
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
Connect the five points with a smooth curve and
graph one complete cycle of the given function
49. The equation
y = −3cos
2 x −
π is of the form
π
2 50. The equation y = −4 cos 2 x −
is of the form 2
y = A cos(Bx − C ) with A = –3, and y = A cos(Bx − C ) with A = –4, and B = 2, and
B = 2, and C = π
. Thus, the amplitude is 2
C = π
. Thus, the amplitude is 2
A = −4
= 4 . The
A = −3 = 3 . The period is 2π
= 2π
= π . The period is
2π =
2π = π . The phase shift is
B 2 B 2 π π
phase shift is C
= 2 = π
⋅ 1
= π
. C
= 2 = π
⋅ 1
= π
. The quarter-period is π
. TheB 2 2 2 4 B 2 2 2 4 4
The quarter-period is π
. The cycle begins at x = π
. cycle begins at x =
π . Add quarter-periods to
4 4 4
Add quarter-periods to generate x-values for the key
points.
x = π 4
x = π
+ π
= π
4 4 2
x = π
+ π
= 3π
2 4 4
generate x-values for the key points.
x = π 4
x = π
+ π
= π
4 4 2
x = π
+ π
= 3π
2 4 4
x = 3π
+ π
= π
x = 3π
+ π
= π 4 4 π 5π4 4
x = π + π
= 5π
4 4
x = π + = 4 4
Evaluate the function at each value of x.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
87 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 87
x
coordinates
π
4
π , − 4
4
2
2
3π
4
4
π
(π , 0)
5π
4
5π , − 4
4
x coordinates
–4 (–4, 2)
15 −
4
15 − , 0 4
7 −
2
−
7 , − 2
2
− 13 4
−
13 , 0
4
–3 (–3, 2)
= − = −
= −
= −
= −
= −
π π
, 0
3π
, 4
Connect the five key points with a smooth curve
and graph one complete cycle of the given function.
52.
Connect the five points with a smooth curve and
graph one complete cycle of the given function
y = 3 cos(2π x + 4π ) = 3 cos(2π x − (−4π ))51. y = 2 cos(2π x + 8π ) = 2 cos(2π x − (−8π ))
The equation y = 3cos(2π x − (−4π )) is of the formThe equation y = 2 cos(2π x − (−8π )) is of the form
y = A cos(Bx − C ) with A = 3, and B = 2π , andy = A cos(Bx − C ) with A = 2, B = 2π , and
C = −4π . Thus, the amplitude is A = 3 = 3 . TheC = −8π . Thus, the amplitude is A = 2 = 2 . The
2π 2π
period is 2π
= 2π
= 1 . The phase shift is period is = = 1 . The phase shift is
B 2πB 2π C −4π 1
C =
−8π = −4 . The quarter-period is
1 . The cycle
= B 2π
= −2 . The quarter-period is . The cycle 4
B 2π 4 begins at x = –2. Add quarter-periods to generate x-begins at x = –4. Add quarter-periods to generate x- values for the key points.
x = −4
x = −4 + 1
= − 15
values for the key points. x = −2
x = −2 + 1
= − 7
4 44 4
x 15
+ 1
= − 7
x 7
+ 1
= − 3
4 4 24 4 2
x 7
+ 1
= − 13
2 4 4
x 13
+ 1
= −3
x 3
+ 1
= − 5
2 4 4
x 5
+ 1
= −1 4 4
4 4
Evaluate the function at each value of x.
Evaluate the function at each value of x.
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88 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 88
x coordinates
–2 (–2, 3)
− 7 4
−
7 , 0
4
− 3 2
−
3 , − 3
2
− 5 4
−
5π , 0
4
–1 (–1, 3)
x y = sin x − 2 coordinates
0 y = sin 0 − 2 = 0 − 2 = −2 (0, –2)
π
2 y = sin
π − 2 = 1 − 2 = −1
2
π , −1
π y = sin π − 2 = 0 − 2 = −2 (π , − 2)
3π
2 y = sin
3π − 2 = −1 − 2 = −3
2
3π , − 3
2π y = sin 2π − 2 = 0 − 2 = −2 (2π , − 2)
x y = sin x + 2 coordinates
0 y = sin 0 + 2 = 0 + 2 = 2 (0, 2)
π
2 y = sin
π + 2 = 1 + 2 = 3
2
π , 3
2
π y = sin π + 2 = 0 + 2 = 2 (π , 2)
3π
2 y = sin
3π + 2 = −1 + 2 = 1
2
3π , 1
2
2π y = sin 2π + 2 = 0 + 2 = 2 (2π , 2)
By connecting the points with a smooth curve we
obtain one period of the graph.
54. The graph of y = sin x − 2 is the graph of y = sin x
Connect the five key points with a smooth curve and
graph one complete cycle of the given function. shifted 2 units downward. The period for both
2π πfunctions is 2π . The quarter-period is
4 or .
2
The cycle begins at x = 0. Add quarter-periods to
generate x-values for the key points. x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
53. The graph of
y = sin x + 2 is the graph of
y = sin x x = π +
π =
3π 2 2
shifted up 2 units upward. The period for both
functions is 2π . The quarter-period is 2π
or π
.
x = 3π
+ π
= 2π 2 2
4 2 Evaluate the function at each value of x.
The cycle begins at x = 0. Add quarter-periods to
generate x-values for the key points.
x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
x = π + π
= 3π
2 2
x = 3π
+ π
= 2π
4
2 2
Evaluate the function at each value of x.
2
By connecting the points with a smooth curve we
obtain one period of the graph.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
89 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 89
x y = cos x − 3 coordinates
0 y = cos 0 − 3 = 1 − 3 = −2
(0, –2)
π
2 y = cos
π − 3
2 = 0 − 3 = −3
π , − 3
π y = cos π − 3 = −1 − 3 = −4
(π , − 4)
3π
2
3π y = cos − 3
2 = 0 − 3 = −3
3π , − 3 2
2π y = cos 2π − 3 = 1 − 3 = −2
(2π , − 2)
x y = cos x + 3 coordinates
0 y = cos 0 + 3 = 1 + 3 = 4 (0, 4)
π
2 y = cos
π + 3 = 0 + 3 = 3
2
π , 3
π y = cos π + 3 = −1 + 3 = 2 (π , 2)
3π
2
3π y = cos + 3 = 0 + 3 = 3
2
3π , 3 2
2π y = cos 2π + 3 = 1 + 3 = 4 (2π , 4)
55. The graph of y = cos x − 3 is the graph of y = cos x 56. The graph of y = cos x + 3 is the graph of y = cos x
shifted 3 units downward. The period for both
functions is 2π . The quarter-period is 2π
or π
.
shifted 3 units upward. The period for both functions
is 2π . The quarter-period is 2π
or π
. The cycle4 2 4 2
The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points. x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
x = π + π
= 3π
2 2
x = 3π
+ π
= 2π 2 2
Evaluate the function at each value of x.
begins at x = 0. Add quarter-periods to generate x- values for the key points. x = 0
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
x = π + π
= 3π
2 2
x = 3π
+ π
= 2π 2 2
Evaluate the function at each value of x.
2
2
By connecting the points with a smooth curve we
obtain one period of the graph.
By connecting the points with a smooth curve we obtain one period of the graph.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
90 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 90
x 1 y = 2sin x + 1
2
coordinates
0 1 y = 2sin
⋅ 0
+ 1 2
= 2sin 0 + 1 = 2 ⋅ 0 + 1 = 0 + 1 = 1
(0, 1)
π 1 y = 2sin ⋅π + 1
2
= 2sin π
+ 1 2
= 2 ⋅1 + 1 = 2 + 1 = 3
(π , 3)
2π 1 y = 2sin ⋅ 2π + 1
2 = 2 sin π + 1 = 2 ⋅ 0 + 1 = 0 + 1 = 1
(2π , 1)
3π y = 2sin
1 ⋅ 3π
+ 1
2
= 2sin 3π
+ 1 2
= 2 ⋅ (−1) + 1 = −2 + 1 = −1
(3π , − 1)
4π 1 y = 2sin ⋅ 4π + 1
2 = 2sin 2π + 1 = 2 ⋅ 0 + 1 = 0 + 1 = 1
(4π , 1)
57. The graph of y = 2sin 1 x + 1 is the graph 2
58. The graph of y = 2 cos 1
x + 1 is the graph of
of y = 2sin 1 x shifted one unit upward. The 2
y = 2 cos 1
x
2
shifted one unit upward. The amplitudeamplitude for both functions is 2 = 2 . The period 2
for both functions is 2π
= 2π ⋅ 2 = 4π . The quarter- 1
for both functions is 2
2π = 2 . The period for both
2
period is 4π
= π . The cycle begins at x = 0. Add
functions is 1
= 2π ⋅ 2 = 4π . The quarter-period is 2
4 quarter-periods to generate x-values for the key points. x = 0 x = 0 + π = π x = π + π = 2π x = 2π + π = 3π x = 3π + π = 4π Evaluate the function at each value of x.
4π = π . The cycle begins at x = 0. Add quarter-
4 periods to generate x-values for the key points. x = 0 x = 0 + π = π x = π + π = 2π x = 2π + π = 3π x = 3π + π = 4π Evaluate the function at each value of x.
x 1
y = 2 cos 2
x + 1
coordinates
0 1
y = 2 cos ⋅ 0
+ 1
2 = 2 cos 0 + 1 = 2 ⋅1 + 1 = 2 + 1 = 3
(0, 3)
π 1 y = 2 cos
2 ⋅π + 1
= 2 cos π
+ 1 2
= 2 ⋅ 0 + 1 = 0 + 1 = 1
(π , 1)
2π 1 y = 2 cos
2 ⋅ 2π + 1
= 2 cos π + 1 = 2 ⋅ (−1) + 1 = −2 + 1 = −1
(2π , −1)
3π 1 y = 2 cos
2 ⋅ 3π + 1
= 2 ⋅ 0 + 1 = 0 + 1 = 1
(3π , 1)
4π 1 y = 2 cos
2 ⋅ 4π + 1
= 2 cos 2π + 1 = 2 ⋅1 + 1 = 2 + 1 = 3
(4π , 3)
By connecting the points with a smooth curve we
obtain one period of the graph.
By connecting the points with a smooth curve we
obtain one period of the graph.
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91 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 91
x y = −3cos 2π x + 2 coordinates
0 y = −3cos(2π ⋅ 0) + 2 = −3cos 0 + 2 = −3 ⋅1 + 2 = −3 + 2 = −1
(0, –1)
1
4 y = −3cos
2π ⋅
1 + 2
= −3cos π
+ 2 2
= −3 ⋅ 0 + 2 = 0 + 2 = 2
1 , 2
1
2 y = −3cos
2π ⋅
1 + 2
= −3cos π + 2 = −3 ⋅ (−1) + 2 = 3 + 2 = 5
1 , 5
3
4 y = −3cos
2π ⋅
3 + 2
= −3cos 3π
+ 2 2
= −3 ⋅ 0 + 2 = 0 + 2 = 2
3 , 2
1 y = −3cos(2π ⋅1) + 2 = −3cos 2π + 2 = −3 ⋅1 + 2 = −3 + 2 = −1
(1, –1)
x y = −3sin 2π x + 2 coordinates
0 y = −3sin(2π ⋅ 0) + 2 = −3sin 0 + 2 = −3 ⋅ 0 + 2 = 0 + 2 = 2
(0, 2)
1
4
1 y = −3sin 2π ⋅
4 + 2
= −3sin π
+ 2 2
= −3 ⋅1 + 2 = −3 + 2 = −1
1 4
, − 1
1
2
1 y = −3sin 2π ⋅ + 2
2 = −3sin π + 2 = −3 ⋅ 0 + 2 = 0 + 2 = 2
1 , 2
2
3
4 y = −3sin 2π ⋅
3 + 2
4
= −3sin 3π
+ 2 2
= −3 ⋅ (−1) + 2 = 3 + 2 = 5
, 5
4
1 y = −3sin(2π ⋅1) + 2 = −3sin 2π + 2 = −3 ⋅ 0 + 2 = 0 + 2 = 2
(1, 2)
59. The graph of y = −3cos 2π x + 2 is the graph of 60. The graph of y = −3sin 2π x + 2 is the graph of
y = −3cos 2π x shifted 2 units upward. The y = −3sin 2π x shifted two units upward. The
amplitude for both functions is −3 = 3 . The period amplitude for both functions is A = −3 = 3 . The
for both functions is 2π
= 1 . The quarter-period is 2π
1 . The cycle begins at x = 0. Add quarter-periods to
4 generate x-values for the key points.
x = 0
x = 0 + 1
= 1
4 4
x = 1
+ 1
= 1
4 4 2
x = 1
+ 1
= 3
2 4 4
x = 3
+ 1
= 1 4 4
Evaluate the function at each value of x.
period for both functions is 2π
= 1 . The quarter- 2π
period is 1
. The cycle begins at x = 0. Add quarter– 4
periods to generate x-values for the key points. x = 0
x = 0 + 1
= 1
4 4
x = 1
+ 1
= 1
4 4 2
x = 1
+ 1
= 3
2 4 4
x = 3
+ 1
= 1 4 4
Evaluate the function at each value of x.
4
4
2
2
3
4 4
By connecting the points with a smooth curve we
obtain one period of the graph.
By connecting the points with a smooth curve we
obtain one period of the graph.
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92 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 92
61. Select several values of x over the interval.
x
0 π
4
π
2
3π
4 π
5π
4
3π
2
7π
4 2π
y1 = 2 cos x 2 1.4 0 −1.4 −2 −1.4 0 1.4 2
y2 = sin x 0 0.7 1 0.7 0 −0.7 −1 −0.7 0
y = 2 cos x + sin x 2 2.1 1 −0.7 −2 −2.1 −1 0.7 2
62. Select several values of x over the interval.
x
0 π
4
π
2
3π
4 π
5π
4
3π
2
7π
4 2π
y1 = 3 cos x 3 2.1 0 −2.1 −3 −2.1 0 2.1 3
y2 = sin x 0 0.7 1 0.7 0 −0.7 −1 −0.7 0
y = 3 cos x + sin x 3 2.8 1 −1.4 −3 −2.8 −1 1.4 3
63. Select several values of x over the interval.
x
0 π
4
π
2
3π
4 π
5π
4
3π
2
7π
4 2π
y1 = sin x 0 0.7 1 0.7 0 −0.7 −1 −0.7 0
y2 = sin 2x 0 1 0 −1 0 1 0 −1 0
y = sin x + sin 2 x 0 1.7 1 −0.3 0 0.3 −1 −1.7 0
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93 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 93
64. Select several values of x over the interval.
x
0 π
4
π
2
3π
4 π
5π
4
3π
2
7π
4 2π
y1 = cos x 1 0.7 0 −0.7 −1 −0.7 0 0.7 1
y2 = cos 2 x 1 0 −1 0 1 0 −1 0 1
y = cos x + cos 2x 2 0.7 −1 −0.7 0 −0.7 −1 0.7 2
65. Select several values of x over the interval.
x
0 π
4
π
2
3π
4 π
5π
4
3π
2
7π
4 2π
y1 = sin x 0 0.7 1 0.7 0 −0.7 −1 −0.7 0
y2 = cos 2 x 1 0 −1 0 1 0 −1 0 1
y = sin x + cos 2 x 1 0.7 0 0.7 1 −0.7 −2 −0.7 1
66. Select several values of x over the interval.
x
0 π
4
π
2
3π
4 π
5π
4
3π
2
7π
4 2π
y1 = cos x 1 0.7 0 −0.7 −1 −0.7 0 0.7 1
y2 = sin 2 x 0 1 0 −1 0 1 0 −1 0
y = cos x + sin 2x 1 1.7 0 −1.7 −1 0.3 0 −0.3 1
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94 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 94
67. Select several values of x over the interval.
x
0 1
2
1 3
2
2 5
2
3 7
2
4
y1 = sin π x 0 1 0 −1 0 1 0 −1 0
π y2 = cos
2 x
1
0.7
0
−0.7
−1
−0.7
0
0.7
1
y = sin π x + cos π
x 2
1
1.7
0
−1.7
−1
0.3
0
−0.3
1
68. Select several values of x over the interval.
x
0 1
2
1 3
2
2 5
2
3 7
2
4
y1 = cos π x 1 0 −1 0 1 0 −1 0 1
π y2 = sin
2 x
0
0.7
1
0.7
0
−0.7
−1
−0.7
0
y = cos π x + sin π
x 2
1
0.7
0
0.7
1
−0.7
−2
−0.7
1
69. Using y = A cos Bx the amplitude is 3 and A = 3 , The period is 4π and thus
B = 2π
= 2π
= 1
period 4π 2
y = A cos Bx
y = 3cos 1
x 2
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
95 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 95
70. Using y = A sin Bx the amplitude is 3 and A = 3 , The 76.
period is 4π and thus
B = 2π
= 2π
= 1
period 4π 2
y = A sin Bx
y = 3sin 1
x 2
71. Using y = A sin Bx the amplitude is 2 and A = −2 ,
The period is π and thus
B = 2π
= 2π
= 2
77.
period π
y = A sin Bx y = −2sin 2 x
72. Using y = A cos Bx the amplitude is 2 and A = −2 ,
The period is 4π and thus
B = 2π
= 2π
= 2
78.
period π
y = A cos Bx y = −2 cos 2x
73. Using y = A sin Bx the amplitude is 2 and A = 2 , The
period is 4 and thus
B = 2π
= 2π
= π
period 4 2
y = A sin Bx
y = 2sin π
x
79.
2
74. Using y = A cos Bx the amplitude is 2 and A = 2 ,
The period is 4 and thus
B = 2π
= 2π
= π
period 4 2
y = A cos Bx
y = 2 cos π
x
80.
2
75.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
96 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 96
81.
82.
92. The information gives the five key points of the graph.
(0, 23) corresponds to Noon,
(3, 38) corresponds to 3 P.M., (6, 53) corresponds to 6 P.M., (9, 38) corresponds to 9 P.M., (12, 23) corresponds to Midnight. By connecting the five key points with a smooth curve
we graph information from noon to midnight. Extend
the graph one cycle to the right to graph the information
for 0 ≤ x ≤ 24.
83. The period of the physical cycle is 33 days.
93. The function y = 3 sin 2π
( x − 79) + 12 is of the form 365
84. The period of the emotional cycle is 28 days. C 2π
85. The period of the intellectual cycle is 23 days. y = A sin B x −
B + D with A = 3 and B =
365 .
86. In the month of February, the physical cycle is at a
minimum on February 18. Thus, the author should
a. The amplitude is A = 3 = 3 .
not run in a marathon on February 18. b. The period is 2π =
2π = 2π ⋅
365 = 365 .
87. In the month of March, March 21 would be the best
day to meet an on-line friend for the first time,
because the emotional cycle is at a maximum.
88. In the month of February, the intellectual cycle is at a
maximum on February 11. Thus, the author should
begin writing the on February 11.
89. Answers may vary.
90. Answers may vary.
91. The information gives the five key point of the graph.
(0, 14) corresponds to June,
(3, 12) corresponds to September, (6, 10) corresponds to December, (9, 12) corresponds to March, (12, 14) corresponds to June By connecting the five key points with a smooth
curve we graph the information from June of one
year to June of the following year.
B 2π 2π 365
c. The longest day of the year will have the most
hours of daylight. This occurs when the sine
function equals 1.
y = 3 sin 2π
( x − 79) + 12 365
y = 3(1) + 12 y = 15
There will be 15 hours of daylight.
d. The shortest day of the year will have the least
hours of daylight. This occurs when the sine
function equals –1.
y = 3 sin 2π
( x − 79) + 12 365
y = 3(−1) + 12 y = 9
There will be 9 hours of daylight.
e. The amplitude is 3. The period is 365. The
phase shift is C
= 79 . The quarter-period is B
365 = 91.25 . The cycle begins at x = 79. Add
4 quarter-periods to find the x-values of the key points.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
97 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 97
3
x = 79 x = 79 + 91.25 = 170.25 x = 170.25 + 91.25 = 261.5 x = 261.5 + 91.25 = 352.75 x = 352.75 + 91.25 = 444
Because we are graphing for 0 ≤ x ≤ 365 , we
will evaluate the function for the first four x-
values along with x = 0 and x = 365. Using a
calculator we have the following points.
(0, 9.1) (79, 12) (170.25, 15) (261.5, 12) (352.75, 9) (365, 9.1)
By connecting the points with a smooth curve
we obtain one period of the graph, starting on
January 1.
The highest average monthly temperature is 56° in
July.
95. Because the depth of the water ranges from a
minimum of 6 feet to a maximum of 12 feet, the
curve oscillates about the middle value, 9 feet. Thus,
D = 9. The maximum depth of the water is 3 feet
above 9 feet. Thus, A = 3. The graph shows that one
complete cycle occurs in 12-0, or 12 hours. The
period is 12. Thus,
12 = 2π
94. The function
y = 16 sin π
x − 2π
+ 40
is in the B 6 3
12B = 2π
form
y = A sin(Bx − C ) + D with A = 16,
B = π
, 6
B = 2π
= π
12 6
and C = 2π
. The amplitude is 3
A = 16
= 16 . The Substitute these values into y = A cos Bx + D . The
π x
period is 2π =
2π = 2π ⋅
6 = 12 . The phase shift is
depth of the water is modeled by y = 3cos
6 + 9 .
B π π 6
2πC 2π 6 12
= = ⋅ = 4 . The quarter-period is = 3 .
96. Because the depth of the water ranges from aB π 3 π 6
4 minimum of 3 feet to a maximum of 5 feet, the curve
The cycle begins at x = 4. Add quarter-periods to find the x-values for the key points.
x = 4 x = 4 + 3 = 7 x = 7 + 3 = 10 x = 10 + 3 = 13
oscillates about the middle value, 4 feet. Thus, D = 4.
The maximum depth of the water is 1 foot above 4
feet. Thus, A = 1. The graph shows that one complete
cycle occurs in 12–0, or 12 hours. The period is 12.
Thus,
2πx = 13 + 3 = 16 12 =
BBecause we are graphing for 1 ≤ x ≤ 12 , we will
evaluate the function for the three x-values between 1
and 12, along with x = 1 and x = 12. Using a
calculator we have the following points.
(1, 24) (4, 40) (7, 56) (10, 40) (12, 26.1) By connecting the points with a smooth curve we
12B = 2π
B = 2π
= π
12 6 Substitute these values into
y = A cos Bx + D . The
π xobtain the graph for 1 ≤ x ≤ 12 .
depth of the water is modeled by y = cos 6
+ 4 .
97. – 110. Answers may vary.
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
98 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 98
111. The function y = 3sin(2 x + π ) = 3 sin(2 x − (−π )) is of We choose −10 ≤ x ≤ 30 , and −1 ≤ y ≤ 1 for our
the form y = A sin(Bx − C ) with A = 3, B = 2, and graph.
C = −π . The amplitude is A = 3 = 3 . The
period is 2π
= 2π
= π . The cycle begins at
B 2
x = C
= −π
= − π
. We choose − π
≤ x ≤ 3π
, and
B 2 2 2 2 −4 ≤ y ≤ 4 for our graph.
114. The function y = 3sin(2 x − π ) + 5 is of the form
y = A cos( Bx − C ) + D with A = 3, B = 2, C = π , and
D = 5. The amplitude is A = 3 = 3 . The period is
2π =
2π = π . The cycle begins at x =
C = π
. Because
B 2 B 2 D = 5, the graph has a vertical shift 5 units upward. We
112. The function
y = −2 cos
2π x −
π is of the form
choose π ≤ x ≤
5π , and 0 ≤ y ≤ 10 for our graph.
2
2 2
y = A cos(Bx − C ) with A = –2, B = 2π , and C = π
. 2
The amplitude is A = −2 = 2 . The period is
2π =
2π = 1 . The cycle begins at
B 2π π
x = C
= 2 = π
⋅ 1
= 1
. We choose 1
≤ x ≤ 9
,B 2π 2 2π 4 4 4
and −3 ≤ y ≤ 3 for our graph.
115.
The graphs appear to be the same from − π
to π
.
113. The function
y = 0.2sin π
x + π = 0.2sin
π x − (−π )
is of the
2 2
116.
10
10
form
y = A sin(Bx − C ) with A = 0.2, B = π
, and 10
C = −π . The amplitude is A = 0.2 = 0.2 . The
period is 2π =
2π = 2π ⋅
10 = 20 . The cycle begins
B π π 10
π πat x =
C =
−π = −π ⋅
10 = −10 .
The graphs appear to be the same from − to . 2 2
B π π 10
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Section 2.1 Graphs of Sine and Cosine Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
99 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 99
117. 125. a. Since A = 3 and D = −2, the maximum will
occur at 3 − 2 = 1 and the minimum will occur
at −3 − 2 = −5 . Thus the range is [−5,1]
Viewing rectangle: − π
, 23π
, π
by [−5,1,1] 6 6 6
b. Since A = 1 and D = −2, the maximum will
The graph is similar to
y = sin x , except the occur at 1 − 2 = −1 and the minimum will occur
118.
amplitude is greater and the curve is less smooth. at −1 − 2 = −3 . Thus the range is [−3, −1]
Viewing rectangle: − π
, 7π
, π
by [−3, −1,1] 6 6 6
126. A = π
B = 2π
= 2π
= 2π period 1
C C = = −2
B 2π
The graph is very similar to
smooth.
119. a. see part c.
y = sin x , except not C = −4π
y = A cos( Bx − C ) y = π cos(2π x + 4π ) or y = π cos [2π ( x + 2)]
b. y = 22.61sin(0.50 x − 2.04) + 57.17
127.
y = sin 2 x = 1
− 1
cos 2 xc. graph for parts a and c: 2 2
120. Answers may vary.
121. makes sense
122. does not make sense; Explanations will vary. Sample
explanation: It may be easier to start at the highest
point.
123. makes sense
124. makes sense
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100 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 100
π
128. y = cos2
x = 1
+ 1
cos 2 x 2 2
132. 4π
lies in quadrant III. The reference angle is 3
θ ′ = 4π
− π = 4π
− 3π
= π
. 3 3 3 3
tan π
= 3 3
Because the tangent is positive in quadrant III,
tan 5π
= + tan π
= 3 4 4
133. − π
< x + π
< π
2 4 2
− π
− π
< x + π
− π
< π
− π
129. Answers may vary. 2 4
2 π 4 4 2 4
π− π
− < x < 2
− π
130. a. cos 47
0 sec 47
0 = cos 470 1
= 1 4 4 4 4
3π πcos 47
0
− < x < 4 4
b. sin 2 θ + cos2 θ = 1
x − 3π
< x < π
or −
3π , π
4 4
4 4
sin 2 π + cos2 π
= 1
5 5 3π π 2π π
131. Use the Pythagorean Theorem,
find b.
c2 = a2 + b2
, to
134.
− + − − 4 4 = 4 = 2 = −
2 2 2 4
a2 + b2 = c2
12 + b2 = 22
1 + b2 = 4
b2 = 3
135. a.
b = 3 = 3
Note that side a is opposite θ and side b is adjacent
to θ .
sin θ = opposite
= 1
hypotenuse 2
cosθ = adjacent
= 3
hypotenuse 2
tan θ = opposite
= 1
= 3
b. The reciprocal function is undefined.
adjacent 3 3
cscθ = hypotenuse
= 2
= 2 opposite 1
secθ = hypotenuse
= 2
= 2 3
adjacent 3 3
cot θ = adjacent
= 3
= 3 opposite 1
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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
2
= −
= −
Section 2.2 y = tan
x −
π from 0 to π . Continue the pattern
Check Point Exercises
1. Solve the equations 2x π
and 2x = π
and extend the graph another full period to the right.
2 2
x= − π x =
π
4 4
Thus, two consecutive asymptotes occur at x π 4
and x = π
. Midway between these asymptotes is 4
x = 0. An x-intercept is 0 and the graph passesthrough (0, 0). Because the coefficient of the tangent
is 3, the points on the graph midway between an x-
intercept and the asymptotes have y-coordinates of
3. Solve the equations
π π–3 and 3. Use the two asymptotes, the x-intercept, and the points midway between to graph one period
x = 0 and 2
x = π 2
π
of y = 3tan 2x from − π 4
to π
. In order to graph for 4
x = 0 x = π 2
x = 2− π
< x < 3π
, Continue the pattern and extend the 4 4
graph another full period to the right.
Two consecutive asymptotes occur at x = 0 and x = 2.
Midway between x = 0 and x = 2 is x = 1. An x-
intercept is 1 and the graph passes through (1, 0).
Because the coefficient of the cotangent is 1
, the 2
points on the graph midway between an x-intercept
and the asymptotes have y-coordinates of − 1 2
and 1
. 2
Use the two consecutive asymptotes, x = 0 and x = 2,
to graph one full period of y = 1
cot π
x . The curve 2 2
2. Solve the equations is repeated along the x-axis one full period as shown.
x − π
= − π
and x − π
= π
2 2 2 2
x = π
− π x =
π + π
2 2 2 2
x = 0 x = π
Thus, two consecutive asymptotes occur at
x = 0 and x = π .
x-intercept = 0 + π
= π
2 2
An x-intercept is π
and the graph passes through 2
π , 0 . Because the coefficient of the tangent is 1,
2
the points on the graph midway between an x-
intercept and the asymptotes have y-coordinates of –1
and 1. Use the two consecutive asymptotes,
x = 0 and x = π , to graph one full period of
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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
C
4. The x-intercepts of
y = sin x +
π correspond to
4.
− π
, 3π
;
− π
; 3π
4
vertical asymptotes of
y = csc x +
π .
4 4 4 4
4
5. 3sin 2 x
6. y = 2 cos π x
7. false
8. true
5. Graph the reciprocal cosine function,
y = 2 cos 2x .
Exercise Set 2.2
πThe equation is of the form y = A cos Bx with A = 2 1. The graph has an asymptote at x = − .
2and B = 2. C π πamplitude: A = 2 = 2 The phase shift, , from
B 2 to −
2 is −π units.
period: 2π
= 2π
= π Thus, C
= C
= −πB 2 B 1
Use quarter-periods, π
, to find x-values for the five 4
key points. Starting with x = 0, the x-values are
C = −π The function with C = −π
is y = tan( x + π ) .
2. The graph has an asymptote at x = 0 .
0, π
, π
, 3π
, and π . Evaluating the function at each 4 2 4
value of x, the key points are
The phase shift,
C , from
π B 2
to 0 is − π
units. Thus, 2
(0, 2), π
, 0 ,
π , −2
, 3π
, 0 , (π , 2) . In order to C C π
= = − 4
2
4
B 1 2 π
= − graph for −
3π ≤ x ≤
3π , Use the first four points
4 4 2
The function with C π
is y = tan x +
π .
and extend the graph − 3π
units to the left. Use the = −
2
2
4 graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts,
3. The graph has an asymptote at x = π .
πand use them as guides to graph y = 2sec 2x . π = + C
2
C = π 2
The function is
y = − tan x −
π . 2
4. The graph has an asymptote at π
. 2
There is no phase shift. Thus, C
= C
= 0
Concept and Vocabulary Check 2.2 B 1
1. − π
, π
;
− π
; π
The function with C = 0 is
C = 0 y = − tan x .
4 4
4 4
2. (0, π ) ; 0; π
3. (0, 2); 0; 2
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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
= −
= −
= −
5. Solve the equations x π
and x = π Continue the pattern and extend the graph another
full period to the right.4 2 4 2
x = − π
4
x = π
4 2
2
x = −2π x = 2π
Thus, two consecutive asymptotes occur at x = −2π and x = 2π .
x-intercept = −2π + 2π
= 0
= 0 2 2
An x-intercept is 0 and the graph passes through (0,
0). Because the coefficient of the tangent is 3, the
7. Solve the equations 2 x
π and 2 x
π
points on the graph midway between an x-intercept
and the asymptotes have y-coordinates of –3 and 3.
= − = 2 2
− π π
Use the two consecutive asymptotes, x = −2π and x = 2 x = 2
x = 2π , to graph one full period of
y = 3tan x
2 2
from π π
−2π 4
to 2π . x = − x =
4 4
Continue the pattern and extend the graph another full period to the right.
Thus, two consecutive asymptotes occur at x π 4
and x = π
. 4
− π + π 0
x-intercept = 4 4 = = 0 2 2
An x-intercept is 0 and the graph passes through (0,
0). Because the coefficient of the tangent is 1
, the 2
points on the graph midway between an x-intercept
6. Solve the equations and the asymptotes have y-coordinates of − 1 and
1 .
x π and
x π 2 2= − =
4 2 4 2 Use the two consecutive asymptotes, x π
and
x = − π
4
x = π
4 4
2
2
1 x = π
, to graph one full period of y =
tan 2x fromx = −2π x = 2π 4 2
Thus, two consecutive asymptotes occur at x = −2π and x = 2π .
− π
to π
. Continue the pattern and extend the graph 4 4
x-intercept =
−2π + 2π =
0 = 0
2 2
another full period to the right.
An x-intercept is 0 and the graph passes through
(0, 0). Because the coefficient of the tangent is 2, the points on the graph midway between an x-intercept and the asymptotes have
y-coordinates of –2 and 2. Use the two consecutive
asymptotes, x = −2π and x = 2π ,
to graph one full period of
2π .
y = 2 tan x 4
from −2π to
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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
= −
= −
8. Solve the equations graph another full period to the right.
π 2 x = − and 2 x =
π
2 2
− π π
x = 2 x = 2
2 2
x π
x π
= − = 4 4
Thus, two consecutive asymptotes occur at x π 4
10. Solve the equations
and x = π
. 1 π 1 π
4
− π + π
x = − 2 2
and x = 2 2
x-intercept = 4 4 = 0
= 0
x = − π
2
x = π
22 2
2
2
An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the tangent is 2, the x = −π x = π
points on the graph midway between an x-intercept and the asymptotes have y-coordinates of –2 and 2.
Thus, two consecutive asymptotes occur at x = −π and x = π .
Use the two consecutive asymptotes, x π
and x-intercept =
−π + π =
0 = 0
x = π
, to graph one full period of 4
4
y = 2 tan 2x
from
2 2 An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the tangent is –3, the
points on the graph midway between an x-intercept
− π
to π
. 4 4
and the asymptotes have y-coordinates of 3 and –3. Use the two consecutive
Continue the pattern and extend the graph another full period to the right.
asymptotes, x = −π and x = π , to graph one full
1period of y = −3 tan
2 x from −π to π . Continue the
9. Solve the equations
pattern and extend the graph another full period to the
right.
1 π x = −
and 1 x =
π
2 2 2 2
x = − π
2
x = π
2 2
2 11. Solve the equations
x = −π x = π π π
Thus, two consecutive asymptotes occur at x = −π x − π = − 2 and x − π =
2and x = π . π π
x = − + π x = + π
x-intercept = −π + π
= 0
= 0 2 2
x = π
2 2
x = 3π
An x-intercept is 0 and the graph passes through (0,
0). Because the coefficient of the tangent is –2, the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of 2 and –2.
2 2
Thus, two consecutive asymptotes occur at x = π 2
3πUse the two consecutive asymptotes, x = −π and and x = .
2
x = π , to graph one full period of y = −2 tan
1 x
π + 3π 4π
from −π
2 to π . Continue the pattern and extend the
x-intercept = 2 2 = 2 = 4π
= π 2 2 4
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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
= −
= −
An x-intercept is π and the graph passes through
(π , 0) . Because the coefficient of the tangent is 1,
the points on the graph midway between an x- intercept and the asymptotes have y-coordinates of –1
and 1. Use the two consecutive asymptotes, x = π
2
Continue the pattern and extend the graph another
full period to the right.
and x = 3π
, to graph one full period of 2
y = tan( x − π ) from π
to 3π
. Continue the pattern 2 2
and extend the graph another full period to the right.
13. There is no phase shift. Thus,
C =
C = 0
B 1 C = 0
Because the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of
–1 and 1, A = –1. The function with C = 0 and A = –1 is y = − cot x .
12. Solve the equations 14. The graph has an asymptote at π
. The phase shift, 2
x − π
= − π
and x − π
= π
C π π4 2 4 2 , from 0 to is units.
x 2π π
x 2π π B 2 2= − + = +
4 4 4 4 C C π
x π
x 3π Thus, = =
= − = B 1 24 4
Thus, two consecutive asymptotes occur at x π
C = π 2
4 The function with C =
π is y = − cot x −
π .
and x = 3π
. 4
2 2
− π
+ 3π 2π
15. The graph has an asymptote at − π
. The phase shift,x-intercept = 4 4 = 4 =
π 2 2 4
C π π
2
C C π, from 0 to − is − units. Thus,
= = −
An x-intercept is π
and the graph passes through 4
π , 0 . Because the coefficient of the tangent is 1,
B 2 2 B 1 2 π
C = − 2
4
= − is .
The function with C π y = cot
x +
π
the points on the graph midway between an x- intercept and the asymptotes have y-coordinates of –1
and 1. Use the two consecutive asymptotes, x π
2
2
16. The graph has an asymptote at −π . The phase shift,
4
and x = 3π
, to graph one full period of
C , from 0 to −π
B is −π units.
4 Thus, C
= C
= −π
y = tan x −
π from 0 to π .
B 1
4 C = −π
The function with C = −π is y = cot( x + π ) .
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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
π
17. Solve the equations x = 0 and x = π . Two
consecutive asymptotes occur at x = 0 and x = π .
x-intercept = 0 + π
= π
2 2
An x-intercept is π
and the graph passes through 2
π , 0 . Because the coefficient of the cotangent is
The curve is repeated along the x-axis one full period
as shown.
2
2, the points on the graph midway between an x- intercept and the asymptotes have y-coordinates of 2 and –2. Use the two consecutive asymptotes, x = 0and x = π , to graph one full period of y = 2 cot x .
19. Solve the equations 2 x = 0 and 2 x = π
The curve is repeated along the x-axis one full period x = 0
x = π 2
as shown. Two consecutive asymptotes occur at x = 0 and
x = π
. 2
0 + π π
x-intercept = 2 = 2 = 2 2 4
An x-intercept is π
and the graph passes through 4
π , 0 . Because the coefficient of the cotangent is
4
18. Solve the equations
1 , the points on the graph midway between an x-
x = 0 and x = π 2
1Two consecutive asymptotes occur at x = 0
and x = π .
x-intercept = 0 + π
= π
intercept and the asymptotes have y-coordinates of 2
and − 1
. Use the two consecutive asymptotes, x = 0 2
2 2
An x-intercept is π
and the graph passes through and x = π
, to graph one full period of 2
y = 1
cot 2x . 2
2
π , 0 . Because the coefficient of the cotangent is
The curve is repeated along the x-axis one full period
as shown. 2
1 , the points on the graph midway between an x-
2
intercept and the asymptotes have y-coordinates of 1 2
and − 1
. Use the two consecutive asymptotes, x = 0 2
and x = π , to graph one full period of y = 1
cot x . 2
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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
π
= −
= −
20. Solve the equations 2 x = 0 and 2 x = π 22. Solve the equations π
x = 0 and π
x = πx = 0 x =
π 2
4 4 x = 0
x = π
π
Two consecutive asymptotes occur at x = 0 and x = π
. 2
0 + π π
x-intercept = 2 = 2 = 2 2 4
4 x = 4
Two consecutive asymptotes occur at x = 0 and
x = 4 .
0 + 4 4
An x-intercept is π
and the graph passes through x-intercept = = = 2 2 2
4
π , 0 . Because the coefficient of the cotangent is 2,
An x-intercept is 2 and the graph passes through
(2, 0 ) . Because the coefficient of the cotangent is – 4
the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of 2 and –2. Use
the two consecutive asymptotes, x = 0 and x = π
, to 2
2, the points on the graph midway between an x- intercept and the asymptotes have y-coordinates of –2
and 2. Use the two consecutive asymptotes, x = 0
and x = 4 , to graph one full period of
graph one full period of
y = 2 cot 2x . The curve is y = −2 cot π
x . The curve is repeated along the x- 4
repeated along the x-axis one full period as shown. axis one full period as shown.
21. Solve the equations π x = 0 and
π x = π
2 2 x = 0
x = π
23. Solve the equations
π ππ x + = 0 and x + = π2 2 2
x = 2 x = 0 −
π x = π − π
Two consecutive asymptotes occur at x = 0 and x = 2. 2 2 π π
x-intercept = 0 + 2
= 2
= 1 2 2
x = − x = 2 2
An x-intercept is 1 and the graph passes through (1, 0).
Because the coefficient of the cotangent is –3, the
points on the graph midway between an x-intercept
and the asymptotes have y-coordinates of –3 and 3.
Use the two consecutive asymptotes, x = 0 and x = 2, to graph one full period of
y = −3cot π
x . The curve is repeated along the x-axis 2
one full period as shown.
Two consecutive asymptotes occur at x π
and 2
x = π
. 2
− π + π 0
x-intercept = 2 2 = = 0 2 2
An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the cotangent is 3, the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of 3 and –3.
Use the two consecutive asymptotes, x π
and
x = π
, to graph one full period of
2
y = 3cot x +
π .
2
2
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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
= −
= −
= −
= −
The curve is repeated along the x-axis one full period 1 xas shown. 25. The x-intercepts of y = − sin
2 2 corresponds to
vertical asymptotes of y 1
csc x
. Draw the 2 2
vertical asymptotes, and use them as a guide to
sketch the graph of y 1
csc x
. 2 2
24. Solve the equations
x + π
= 0 and
x + π
= π4 4
x = 0 − π x = π −
π
4 4
x π
x 3π
= − = 4 4
26. The x-intercepts of y = 3sin 4x correspond to
Two consecutive asymptotes occur at x π
and vertical asymptotes of y = 3csc 4x . Draw the vertical
4 asymptotes, and use them as a guide to sketch the
x = 3π
. 4
− π + 3π
2π
π
graph of y = 3csc 4x .
x-intercept = 4 4 = 4 = 2 2 4
An x-intercept is π
and the graph passes through 4
π , 0 . Because the coefficient of the cotangent is
4
3, the points on the graph midway between an x- intercept and the asymptotes have y-coordinates of 3
and –3. Use the two consecutive asymptotes, x π
27. The x-intercepts of
y = 1
cos 2π x corresponds to 2
4
and x = 3π
, to graph one full period of
vertical asymptotes of y =
1 sec 2π x . Draw the
24
y = 3cot x +
π . The curve is repeated along the x-
vertical asymptotes, and use them as a guide to
1 4
sketch the graph of y =
sec 2π x . 2
axis one full period as shown.
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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
28. The x-intercepts of y = −3cos π
x correspond to 30. Graph the reciprocal sine function, y = 2sin x . The
vertical asymptotes of
2
y = −3sec π
x . Draw the
equation is of the form
B = 1 .
y = A sin Bx with A = 2 and
2 vertical asymptotes, and use them as a guide to
amplitude:
2π A = 2 = 2
2π
sketch the graph of y = −3sec π
x . 2
period: = = 2π B 1
πUse quarter-periods, , to find x-values for the five
2key points. Starting with x = 0, the x-values are
0, π
, π , 3π
, and 2π . Evaluating the function at 2 2
each value of x, the key points are
(0, 0), π
, 2 , (π , 0),
3π , − 2
, and (2π , 0) .
2
2
29. Graph the reciprocal sine function,
y = 3sin x . The Use these key points to graph y = 2sin x from 0 to
equation is of the form
B = 1.
y = A sin Bx with A = 3 and 2π . Extend the graph one cycle to the right. Use the graph to obtain the graph of the reciprocal function.
Draw vertical asymptotes through the x-intercepts,amplitude: A = 3 = 3 and use them as guides to graph y = 2cscx.
period: 2π
= 2π
= 2π B 1
Use quarter-periods, π
, to find x-values for the 2
five key points. Starting with x = 0, the x-values are
0, π
, 2 π ,
3π , and 2π . Evaluating the function at
2each value of x, the key points are (0, 0),
π
, 3 , (π , 0),
3π , − 3
, and (2π , 0) . Use
31. Graph the reciprocal sine function, y = 1
sin x
. The
2
2
2 2
these key points to graph y = 3sin x from 0 to 2π .
equation is of the form y = A sin Bx with A = 1
andExtend the graph one cycle to the right.
Use the graph to obtain the graph of the reciprocal
function. Draw vertical asymptotes through the x-
intercepts, and use them as guides to graph
2
B = 1
. 2
1 1
y = 3csc x . amplitude: A = = 2 2
period: 2π
= 2π
= 2π ⋅ 2 = 4π B 1
2
Use quarter-periods, π , to find x-values for the five
key points. Starting with x = 0, the x-values are 0,
π , 2π , 3π , and 4π . Evaluating the function at each
value of x, the key points are (0, 0),
π ,
1 , (2π , 0),
3π , −
1 , and (4π , 0) . Use these
2
2
key points to graph y = 1
sin x
2 2
from 0 to 4π .
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110 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 110
Extend the graph one cycle to the right. Use the graph 33. Graph the reciprocal cosine function, y = 2 cos x .
to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts, and use
The equation is of the form
and B = 1.
y = A cos Bx with A = 2
them as guides to graph y = 1
csc x
. 2 2
amplitude: A = 2 = 2
period: 2π
= 2π
= 2π B 1
Use quarter-periods, π
, to find x-values for the five 2
key points. Starting with x = 0, the x-values are 0,
π , π ,
3π , 2π . Evaluating the function at each value
2 2
of x, the key points are (0, 2), π
, 0 , (π , − 2),
2
32. Graph the reciprocal sine function,
y = 3
sin x
. The
3π 2 4
, 0 , and (2π , 2) . Use these key points to 2
equation is of the form y = A sin Bx with A = 3 and graph y = 2 cos x from 0 to 2π . Extend the graph
B = 1
. 4
2 one cycle to the right. Use the graph to obtain the
graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts, and use them as
amplitude: A = 3
= 3 guides to graph y = 2sec x .
2 2
period: 2π
= 2π
= 2π ⋅ 4 = 8π B 1
4
Use quarter-periods, 2π , to find x-values for the five
key points. Starting with x = 0, the x-values are
0, 2π , 4π , 6π , and 8π . Evaluating the function at
each value of x, the key points are
(0, 0), 2π ,
3 , (4π , 0),
6π , −
3 , and (8π , 0) .
34. Graph the reciprocal cosine function,
y = 3cos x .
2
2 The equation is of the form y = A cos Bx with A = 3
Use these key points to graph y = 3
sin x
2 4
from 0 to and B = 1 .
amplitude:
A = 3 = 3
8π . Extend the graph one cycle to the right.
period: 2π
= 2π
= 2πUse the graph to obtain the graph of the reciprocal B 1
function. Draw vertical asymptotes through the x-
intercepts, and use them as guides to graph
y = 3
csc x
. 2 4
Use quarter-periods, π
, to find x-values for the five 2
key points. Starting with x = 0, the x-values are
0, π
, π , 3π
, and 2π . Evaluating the function at 2 2
each value of x, the key points are
(0, 3), π
, 0 , (π , − 3), 3π
, 0 , (2π , 3) .
2
2
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111 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 111
Use these key points to graph y = 3cos x from 0 to 36. Graph the reciprocal cosine function, y = cos
x . The
2π . Extend the graph one cycle to the right. Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts,
equation is of the form
1
y = A cos Bx
with
2 A = 1 and
and use them as guides to graph y = 3sec x . B = . 2
amplitude:
A = 1 = 1
period: 2π
= 2π
= 2π ⋅ 2 = 4π B 1
2
35. Graph the reciprocal cosine function,
y = cos x
. The 3
Use quarter-periods, π , to find x-values for the five
key points. Starting with x = 0, the x-values are
0, π , 2π , 3π , and 4π . Evaluating the function at
each value of x, the key points are
(0, 1), (π , 0 ) , (2π , − 1), (3π , 0) , and (4π , 1) .
equation is of the form y = A cos Bx with A = 1 and
Use these key points to graph y = cos x
2
from 0 to
B = 1
. 3
amplitude:
A = 1 = 1
4π . Extend the graph one cycle to the right. Use the
graph to obtain the graph of the reciprocal function.
Draw vertical asymptotes through the x-intercepts,
period: 2π
= 2π
= 2π ⋅ 3 = 6π
and use them as guides to graph y = sec x
.B 1 2
3
Use quarter-periods, 6π
= 3π
, to find x-values for 4 2
the five key points. Starting with x = 0, the x-values
are 0, 3π
, 3π , 9π
, and 6π . Evaluating the function 2 2
at each value of x, the key points are (0, 1), 3π
, 0 ,
2
(3π , −1),
9π , 0 , and (6π , 1) . Use these key
2 37. Graph the reciprocal sine function, y = −2sin π x .
The equation is of the form
y = A sin Bx with A = –2points to graph y = cos
x
3 from 0 to 6π . Extend the
and B = π .
graph one cycle to the right. Use the graph to obtain the graph of the reciprocal function. Draw vertical
amplitude:
2π A = −2 = 2
2πasymptotes through the x-intercepts, and use them as
period: = = 2 B π
guides to graph y = sec x
. 3
Use quarter-periods, 2
= 1
, to find 4 2
x-values for the five key points. Starting with x = 0,
the x-values are 0, 1
, 1, 3
, and 2. Evaluating the 2 2
function at each value of x, the key points are (0, 0),
1 , − 2
, (1, 0),
3 , 2
, and (2, 0) . Use these key
2
2
points to graph
y = −2sin π x from 0 to 2. Extend the
graph one cycle to the right. Use the graph to obtain
the graph of the reciprocal function.
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112 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 112
= −
= −
= −
2 2
2 2
= −
= −
= −
= −
Draw vertical asymptotes through the x-intercepts, 1and use them as guides to graph y = −2 csc π x . 39. Graph the reciprocal cosine function, y = − cos π x .
2The equation is of the form y = A cos Bx with
A 1
and B = π . 2
amplitude: A = − 1
= 1
2 2
period: 2π
= 2π
= 2
B π
38. Graph the reciprocal sine function, y 1
sin π x . 2
Use quarter-periods, 2
= 1
, to find x-values for the 4 2
The equation is of the form y = A sin Bx with five key points. Starting with x = 0, the x-values are
1 3
A 1
and B = π . 0, , 1, , and 2. Evaluating the function at each
2 22
1
amplitude:
A = − 1
= 1 value of x, the key points are 0, −
2 ,
2 2
3
1
1 , 0 , 1,
1 ,
, 0 , 2, − . Use these key
period: 2π
= 2π
= 2
2
2
2 2
B π
1
Use quarter-periods, 2
= 1
, to find x-values for the 4 2
points to graph y = − cos π x from 0 to 2. Extend 2
five key points. Starting with x = 0, the x-values are
0, 1
, 1, 3
, and 2 . Evaluating the function at each 2 2
value of x, the key points are
the graph one cycle to the right. Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts, and use them as guides to graph
1
(0, 0), 1
, − 1
, (1, 0), 3
, 1
, and (2, 0) .
y = − sec π x . 2
Use these key points to graph y 1
sin π x from 0 2
to 2 . Extend the graph one cycle to the right. Use the
graph to obtain the graph of the reciprocal function.
Draw vertical asymptotes through the x-intercepts,
and use them as guides to graph y 1
csc π x . 2
40. Graph the reciprocal cosine function,
y 3
cos π x . 2
The equation is of the form y = A cos Bx with
A 3
and B = π . 2
amplitude: A = − 3
= 3
2 2
period: 2π
= 2π
= 2
B π
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113 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 113
= −
= −
.
Use quarter-periods, 2
= 1
, to find x-values for the Draw vertical asymptotes through the x-intercepts,
4 2 five key points. Starting with x = 0, the x-values are
0, 1
, 1, 3
, and 2 . Evaluating the function at each 2 2
value of x, the key points are
0, −
3 ,
1 , 0
, 1,
3 ,
3 , 0
,
2, −
3 .
and use them as guides to graph y = csc( x − π ) .
2
2
2
2
2
Use these key points to graph y 3
cos π x from 0 2
to 2 . Extend the graph one cycle to the right. Use the 42. Graph the reciprocal sine function,
y = sin
x −
π .
graph to obtain the graph of the reciprocal function.
2
Draw vertical asymptotes through the x-intercepts, The equation is of the form
y = Asin(Bx − C) with
and use them as guides to graph y 3
sec π x . 2
π A = 1 , B = 1 , and C = .
2
amplitude: A = 1 = 1
period: 2π
= 2π
= 2π B 1
π
phase shift: C
= 2 = π
B 1 2
41. Graph the reciprocal sine function,
y = sin(x − π ) .
Use quarter-periods, π
, to find x-values for the five 2
key points. Starting with x = π
, the x-values areThe equation is of the form
= 1, and B = 1, and C = π .
y = Asin(Bx − C) with A 2
π , π ,
3π , 2π , and
5π . Evaluating the function at
amplitude: A = 1 = 1 2 2 2 each value of x, the key points are
period: 2π
= 2π
= 2π
π 3π
5π B 1
, 0 , (π , 1) , , 0 , ( 2π , − 1) , and , 0 . 2
2
2
phase shift: C
= π
= π
B 1 Use these key points to graph
y = sin x −
π from
2 Use quarter-periods,
2π = π
, to find 4 2
π to
5π . Extend the graph one cycle to the right.
x-values for the five key points. Starting with x = π , 2 2
the x-values are π , 3π
, 2π , 2
5π , and 3π .
2
Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-
Evaluating the function at each value of x, the key intercepts, and use them as guides to graph
points are (π , 0), 3π
, 1 , (2π , 0),
y = csc x −
π
2
2
5π , − 1
, (3π , 0) . Use these key points to graph
2
y = sin(x − π ) from π to 3π . Extend the graph one
cycle to the right. Use the graph to obtain the graph
of the reciprocal function.
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114 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 114
2
− π
C
= −
43. Graph the reciprocal cosine function,
y = 2 cos( x + π ) . The equation is of the form
y = A cos(Bx + C ) with A = 2, B = 1, and C = −π .
Evaluating the function at each value of x, the key
points are
− π
, 2 , (0, 0 ) ,
π , − 2
, (π , 0 ) ,
3π , 2
.
amplitude: A = 2 = 2 2 2 2
π
period: 2π
= 2π
= 2π Use these key points to graph y = 2 cos x +
2 from
B 1
− π
to 3π
. Extend the graph one cycle to the right.
phase shift: C
= −π
= −π
B 1 2 2
Use quarter-periods, 2π
= π
, to find x-values for the 4 2
Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x- intercepts, and use them as guides to graph
five key points. Starting with x = −π , the x-values π .
are −π , − π
, 2
0, π
, and π . Evaluating the function 2
y = 2sec x +
at each value of x, the key points are (−π , 2),
, 0 , (0, − 2 ) , π
, 0 , and (π , 2) . Use these
2
2
key points to graph
y = 2 cos( x + π ) from −π
to π .
Extend the graph one cycle to the right. Use the graph
to obtain the graph of the reciprocal function. Draw
vertical asymptotes through the x-intercepts, and use
them as guides to graph y = 2 sec( x + π ) . 45.
44. Graph the reciprocal cosine function,
y = 2 cos x +
π . The equation is of the form
46.
2
y = A cos(Bx + C ) with
π = − .
2
A = 2
and B = 1 , and
amplitude: A = 2 = 2
period: 2π
= 2π
= 2π B 1
π
phase shift: C
= −
2 = − π
B 1 2
Use quarter-periods, π
, to find x-values for the five 2
key points. Starting with x π
, the x-values are 2
− π
, 0, π
, π , and 3π
. 2 2 2
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115 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 115
2
, −
, −
,
47.
52.
53.
y = ( f h ) ( x) = f (h( x)) = 2 sec 2 x −
π
2
48.
π
49.
50.
54. y = ( g h ) ( x) = g (h( x)) = −2 tan 2 x −
55. Use a graphing utility with y1 = tan x and y2 = −1 .
For the window use Xmin = −2π , Xmax = 2π ,
Ymin = −2 , and Ymax = 2 .
5π π x = − ,
3π ,
7π
4 4 4 4
51. x ≈ −3.93, − 0.79, 2.36, 5.50
56. Use a graphing utility with y1 = 1/ tan x and
y2 = −1 .
For the window use Xmin = −2π , Xmax = 2π ,
Ymin = −2 , and Ymax = 2 .
5π π x = − ,
3π ,
7π
4 4 4 4x ≈ −3.93, − 0.79, 2.36, 5.50
57. Use a graphing utility with y1 = 1/ sin x and y2 = 1 .
For the window use Xmin = −2π , Xmax = 2π ,
Ymin = −2 , and Ymax = 2 .
3π π x = −
2 2 x ≈ −4.71, 1.57
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116 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 116
0
= −
= −
58. Use a graphing utility with y1 = 1/ cos x and y2 = 1 . Two consecutive asymptotes occur at x = 0 and
For the window use Xmin = −2π , Xmax = 2π , Ymin = −2 , and Ymax = 2 .
x = −2π , 0, 2π
x = π . Midway between x = 0 and x = π is x = π
. 2
59.
x ≈ −6.28, 0, 6.28
d = 12 tan 2π t
An x-intercept is π
and the graph passes through 2
π , 0 . Because the coefficient of the cotangent is
a. Solve the equations
2
2π t π
and 2π t = π
2, the points on the graph midway between an x-
2 2 − π π
intercept and the asymptotes have y-coordinates of –2 and 2. Use the two consecutive asymptotes, x = 0
t = 2
2π t = 2
2π and x = π , to graph y = 2 cot x for 0 < x < π .1
t = − t = 1
4 4
Thus, two consecutive asymptotes occur at
1 x = −
4 and x =
1 .
4 1 1− +
x-intercept = 4 4 = = 0 2 2
An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the tangent is 12, the points on the graph midway between an
x-intercept and the asymptotes have y-
coordinates of –12 and 12. Use the two
61. Use the function that relates the acute angle with the
hypotenuse and the adjacent leg, the secant function.
sec x = d
10 d = 10 sec x
consecutive asymptotes, x 1 and x =
1 , to Graph the reciprocal cosine function, y = 10 cos x .
4 4
graph one full period of d = 12 tan 2π t . To
The equation is of the form
A = 10 and B = 1.
y = A cos Bx with
graph on [0, 2], continue the pattern and extend the graph to 2. (Do not use the left hand side of
amplitude: A = 10 = 10
the first period of the graph on [0, 2].) period: 2π
= 2π
= 2πB 1
For − π
< x < π
, use the x-values − π
, 0, and π
to2 2 2
2
π
find the key points − π
, 0 , (0, 10), and , 0 . 2
2
Connect these points with a smooth curve, then draw
vertical asymptotes through the x-intercepts, and use
them as guides to graph d = 10 sec x on − π
, π
.
2 2
b. The function is undefined for t = 0.25, 0.75,
1.25, and 1.75.
The beam is shining parallel to the wall at these times.
60. In a right triangle the angle of elevation is one of the
acute angles, the adjacent leg is the distance d, and
the opposite leg is 2 mi. Use the cotangent function.
cot x = d 2
d = 2 cot x
Use the equations x = 0 and x = π .
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Section 2.2 Graphs of Other Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
117 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 117
62. Graphs will vary.
78. period: π
= π
B 4
Graph y = tan 4x for 0 ≤ x ≤ π
. 2
63.
79. period: π
= π
B 2Graph y = cot 2x for 0 ≤ x ≤ π .
64. Graphs will vary.
80. period: π
= π
= π ⋅ 2 = 2πB 1
2
y = x
for 0 ≤ x ≤ 4π .65. – 76. Answers may vary.
77. period: π
= π
= π ⋅ 4 = 4π
Graph cot 2
B 1 4
Graph y = tan x 4
for 0 ≤ x ≤ 8π .
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118 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 118
−1
2
81. period: π
= π
= 1 84. period:
2π =
2π = 2π ⋅
3 = 6
B π B π π 3
Graph y = 1
tan π x for 0 ≤ x ≤ 2 . 2
Graph the functions for 0 ≤ x ≤ 12 .
82. Solve the equations 85. period:
2π =
2π = π
π x + 1 = − π
and π x + 1 = π B 2
2 2 C π
π π
π x = − π x = π
− 1 phase shift: = 6 =
B 2 122 2
−π π
Thus, we include π
≤ x ≤ 25π
in our graph, and− 1 x = 2
− 1 x = 2 12 12
π π graph for 0 ≤ x ≤
5π .
x = −π − 2
2π
x ≈ −0.82
period: π
= π
= 1
x = π − 2 2
2π x ≈ 0.18
B π
Thus, we include −0.82 ≤ x ≤ 1.18 in our graph of
y = 1
tan(π x + 1) , and graph for −0.85 ≤ x ≤ 1.2 . 2
86. period: 2π
= 2π
= 2B π
π
phase shift: C
= 6 = π
⋅ 1
= 1
B π 6 π 6
Thus, we include 1
≤ x ≤ 25
in our graph, and graph 6 6
83. period: 2π
= 2π
= 2π ⋅ 2 = 4π for 0 ≤ x ≤ 9
.B 1
2
Graph the functions for 0 ≤ x ≤ 8π .
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119 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 119
2 2
87. 1 and –1. Thus, A = 1. There is no phase shift. Thus,
C = 0. An equation for this graph is y = cot 3
x . 2
94. The graph has the shape of a secant function.
The reciprocal function has amplitude
period is 8π
. Thus, 2π
= 8π
A = 1 . The
3 B 3The graph shows that carbon dioxide concentration
rises and falls each year, but over all the
concentration increased from 1990 to 2008.
8π B = 6π
B = 6π
= 3
8π 4
There is no phase shift. Thus, C = 0 . An equation for
88. y = sin 1 x
the reciprocal function is
equation for this graph is
y = cos 3
x . Thus, an 4
y = sec 3
x . 4
95. The range shows that A = 2.
Since the period is 3π , the coefficient of x is given
by B where 2π
= 3π B
The graph is oscillating between 1 and –1.
The oscillation is faster as x gets closer to 0.
Explanations may vary.
89. makes sense
2π = 3π
B 3Bπ = 2π
B = 2
3
2 x
90. makes sense Thus, y = 2 csc
3
91. does not make sense; Explanations will vary.
96. The range shows that
A = π .
Sample explanation: To obtain a cosecant graph,
you can use a sine graph.
92. does not make sense; Explanations will vary.
Sample explanation: To model a cyclical temperature, use sine or cosine.
93. The graph has the shape of a cotangent function with
Since the period is 2, the coefficient of x is given by
B where 2π
= 2 B
2π = 2
B 2B = 2π
B = π
consecutive asymptotes at Thus, y = π cscπ x
x = 0 and x = 2π
. The period is 2π
− 0 = 2π
. Thus,
97. a. Since A=1, the range is ( ] [ )3 3 3 −∞, −1 1, ∞
π =
2π Viewing rectangle: −
π ,π ,
7π by [−3, 3,1]
B 3
6 6
2π B = 3π
B = 3π
= 3
b. Since A=3, the range is (−∞, −3] [3, ∞ )2π 2
1 7 The points on the graph midway between an
x-intercept and the asymptotes have y-coordinates of
Viewing rectangle: − , ,1 by [−6, 6,1]
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120 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 120
o
98. y = 2− x sin x
2− x
decreases the amplitude as x gets larger.
Examples may vary.
102. a.
99. The formula s = rθ can only be used when θ is
expressed in radians. Thus, we begin by converting
150° to radians. Multiply by π radians
. 180°
150° = 150° ⋅ π radians
= 150
π
radians
b. yes; Explanations will vary.180°
= 5 π radians
6
180
c. The angle is − π
. 6
Now we can use the formula s = rθ to find the This is represented by the point
− π
, − 1
.length of the arc. The circle’s radius is 8 inches : r = 8 inches. The measure of the central angle in
radians is 5 π : θ =
5 π . The length of the arc
6 6
intercepted by this central angle is s = rθ
= (8 inches) 5 π
6 2
103. a.
6
= 20π 3
inches
b. yes; Explanations will vary.
≈ 20.94 inches.
100. 25π
− 2π ⋅ 4 = 25π
− 8π
c. The angle is 5π
. 6
5π
3
101.
3 3
25π 24π π = − =
3 3 3
tan 35o
= a 120
a = 120 tan 35 a ≈ 120(0.7002) ≈ 84 m
104. a.
This is represented by the point 6
, − 2
.
b. yes; Explanations will vary.
c. The angle is − π
. 3
This is represented by the point − π
, −
3 . 3
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Mid-Chapter 2 Check Point Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
x y 4sin 2x
coordinates
0
4sin(2 0) 4 0
0
(0, 0)
π
4
π 4sin
2
4 4 1
4
π ,4
π
2
4 sin 2 π
4 0
0
π ,0
3π
4
4 sin 2
3π 4(1)
4
3π
, 4
4
π 4sin(2 π ) 4 0
0
(π , 0)
x
1 π y
2 cos
3 x
coordinates
0
1 cos
π 0
1
1 1
2
0, 1
2
3
2
cos 0 0 2 3
3
2 , 0
3
1 cos
π 3
1
1 1
2
3,
1
9
2
1 cos
π 9
1
0 0 2
9 , 0
6
1 π
1
1
2 cos
3 6
2 1
2
6, 1
2
Mid-Chapter 2 Check Point 2. The equation y
1 cos π
x is of the form
1. The equation y 4sin 2x is of the form y Asin Bx 2 3
1 πwith A = 4 and B 2 . Thus, the amplitude is y A cos Bx with A and B . The amplitude
A 4 4 . The period is 2π
2π
π . The
2 3
1 1 2π 2πB 2 is A . The period is
2 2 6 . The B π
quarter-period is π
. The cycle begins at x = 0. Add 4
quarter-periods to generate x-values for the key
points and evaluate the function at each value of x.
3
quarter-period is 6
3
. Add quarter-periods to 4 2
generate x-values for the key points and evaluate the function at each value of x.
4 1 π 3 1
2 2
2
4 2
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
Connect the five points with a smooth curve and
graph one complete cycle of the given function.
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Mid-Chapter 2 Check Point Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
x
y 3sin( x π ) coordinates
π
3sin π π 3sin(0)
3 0
0
π , 0
3π
2
3sin 3π
π 3sin
π
3 1
3
3π
, 3
2π
3sin 2π π 3sin(π )
3 0
0
(2π , 0)
5π
2
3sin 5π
π 3sin
3π
3(1)
3
5π 2
, 3
3π
3sin 3π π 3sin(2π )
3 0
0
3π , 0
x
coordinates
π
8
π , 2
3π
8
3π
8 , 0
5π
8
5π , 2
8
7π
8
7π
8 , 0
9π
8
9π
8 , 2
3. The equation y 3sin( x π ) is of the form π
y Asin( Bx C) with A = 3, B = 1, and C π . The 4. The equation y 2cos
2 x
4 is of the form
amplitude is A 3 3 . The period is y Acos( Bx C) with A = 2, and B = 2, and
2π 2π
2π . The phase shift is C
π
π . The C π
. Thus, the amplitude is
A 2
2 . TheB 1 B 1 4
quarter-period is 2π
π
. The cycle begins at x π . period is 2π
2π
π . The phase shift is4 2
Add quarter-periods to generate x-values for the key
points and evaluate the function at each value of x.
B 2
C π
π 4 .
B 2 8
The quarter-period is π
. The cycle begins at x π
.
4 4 Add quarter-periods to generate x-values for the key points and evaluate the function at each value of x.
2 2 8
2
2 2
Connect the five points with a smooth curve and
graph one complete cycle of the given function.
Connect the five points with a smooth curve and graph one complete cycle of the given function
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Mid-Chapter 2 Check Point Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
x y cos 2 x 1 coordinates
0 y cos 2 0 1 2 (0, 2)
π
4 y cos
2 π
1 1 π
,1
π
2 y cos
2 π
1 0 π
,0
3π
4 y cos
2
3π 1 1
3π ,1
π y cos 2 π 1 2 (π , 2)
5. The graph of y cos 2 x 1 is the graph of y cos 2 x shifted one unit upward. The amplitude for both functions is
1 1 . The period for both functions is 2π
π . The quarter-period is π
. The cycle begins at x = 0. Add quarter-
2 4
periods to generate x-values for the key points and evaluate the function at each value of x.
4 4
2 2
4 4
By connecting the points with a smooth curve we obtain one period of the graph.
6. Select several values of x over the interval.
x
0 π
4
π
2
3π
4 π
5π
4
3π
2
7π
4 2π
y1 2 sin x 0 1.4 2 1.4 0 1.4 2 1.4 0
y2 2 cos x 2 1.4 0 1.4 2 1.4 0 1.4 2
y 2 sin x 2 cos x 2 2.8 2 0 2 2.8 2 0 2
7. Solve the equations
π π x
and π
x π
4 2 4 2
x π 4
x π 4
x 2
2 π 2 π
x 2
Thus, two consecutive asymptotes occur at x 2 and x 2 .
x-intercept = 2 2
0
0 2 2
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Mid-Chapter 2 Check Point Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
An x-intercept is 0 and the graph passes through (0, 0).
Because the coefficient of the tangent is 2, the points
Use quarter-periods, 2 1
, to find x-values for the 4 2
on the graph midway between an x-intercept and the asymptotes have y-coordinates of –2 and 2. Use the two consecutive asymptotes, x 2 and x 2 , to
five key points. Starting with x = 0, the x-values are
0, 1
, 1, 3
, and 2 . Evaluating the function at each 2 2
graph one full period of y 2 tan π
x 4
from 2 to 2 . value of x, the key points are
Continue the pattern and extend the graph another full 0, 2 , 1
, 0 , 1, 2 ,
3 , 0
, 2, 2 .
period to the right. 2 2
Use these key points to graph y 2 cos π x from 0
to 2 . Extend the graph one cycle to the right. Use the
graph to obtain the graph of the reciprocal function.
Draw vertical asymptotes through the x-intercepts,
8. Solve the equations 2 x 0 and 2 x π
and use them as guides to graph y 2sec π x .
x 0 x
π 2
Two consecutive asymptotes occur at x 0 and x π
. 2
0 π π π x-intercept = 2 2
10. Graph the reciprocal sine function,
y 3sin 2π x . The2 2 4
An x-intercept is π
and the graph passes through equation is of the form
B 2π .
y Asin Bx with A 3 and
4
π , 0 . Because the coefficient of the cotangent is 4,
amplitude: A 3 3
4 period: 2π
2π
1
the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of 4 and –4. Use
the two consecutive asymptotes, x 0 and x π
, to 2
B 2π
Use quarter-periods, 1
, to find x-values for the five key 4
points. Starting with x = 0, the x-values are 0, 1
, 1
, 3
, andgraph one full period of y 4 cot 2 x . The curve is 4 2 4
repeated along the x-axis one full period as shown. 1. Evaluating the function at each value of x, the key
points are (0, 0), 1
, 3 ,
1 , 0
,
3 , 3 , and (1, 0) .
4 2 4
Use these key points to graph y 3sin 2π x from 0 to 1 .
Extend the graph one cycle to the right. Use the graph to
obtain the graph of the reciprocal function. Draw vertical
asymptotes through the x-intercepts, and use them as
guides to graph y 3csc 2π x .
9. Graph the reciprocal cosine function, y 2 cos π x .
The equation is of the form
A 2 and B π .
y A cos Bx with
amplitude: A 2 2
period: 2π
2π
2 B π
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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
Scientific Calculator Solution
Function
Mode
Keystrokes Display
(rounded to four places)
cos−1 1
3
Radian
1 ÷ 3 = COS−1
1.2310
tan−1
(−35.85)
Radian
35.85 + TAN−1
−
–1.5429
Graphing Calculator Solution
Function
Mode
Keystrokes Display
(rounded to four places)
cos−1 1
Radian
COS−1
( 1 ÷ 3 ) ENTER
1.2310
tan−1
(−35.85)
Radian
TAN−1 − 35.85 ENTER
–1.5429
2
= −
.
= −
= −
Section 2.3
Check Point Exercises
1. Let θ = sin −1 3 , then sin θ =
3 .
2 2
The only angle in the interval − π
, π
that satisfies sin θ = 3
is π
. Thus, θ = π
, or sin−1 3 = π
. 2 2
2 3 3 2 3
2. Let θ = sin−1 − 2
, then sin θ = − .
2 2
The only angle in the interval − π
, π
that satisfies cosθ = − 2
is − π
. Thus θ π
, or sin−1 − 2
= − π
. 2 2 = −
2 4 4 2 4
3. Let θ = cos−1 − 1
, then cosθ 1
. The only angle in the interval [0, π ] that satisfies cosθ 1
is 2π
. Thus, 2
2 2 3
θ = 2π
, or cos−1 − 1
= 2π
.
3
2
3
4. Let θ = tan−1
(−1) , then tan θ = −1 . The only angle in the interval − π
, π
that satisfies tan θ = −1 is − π
. Thus
2 2
4
π
θ = − 4
or tan−1 θ
π 4
5.
a.
b.
a.
3
b.
6. a. cos (cos−1 0.7) x = 0.7 , x is in [–1,1] so cos(cos
−1 0.7) = 0.7
Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-2nd-Edition-Blitzer
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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
4 x + 1 4
b. sin −1
(sin π )
x = π , x is not in − π
, π
. x is in the domain
9. Let θ = tan −1 x , then tan θ =x= x
. 1
2 2
of sin x, so sin−1
(sin π ) = sin −1
(0) = 0
c. cos (cos−1 π )
x = π , x is not in [–1,1] so cos (cos−1 π ) is not
defined.
7. Let θ = tan −1 3
, then tan θ = 3
. Because tan θ is
Use the Pythagorean Theorem to find the third side, a.
a2
= x2
+ 12
a = x2
+ 1
Use the right triangle to write the algebraic expression. 2
sec ( tan −1 x ) = secθ = =
x2
+ 1positive, θ is in the first quadrant. 1
Concept and Vocabulary Check 2.3
1. − π
≤ x ≤ π
; 2 2
sin −1
x
Use the Pythagorean Theorem to find r.
r 2
= 32
+ 42
= 9 + 16 = 25
2. 0 ≤ x ≤ π ; cos−1
x
r = 25 = 5 3. − π
≤ x ≤ π
; tan −1 x
Use the right triangle to find the exact value. 2 2
sin tan
−1 3 = sin θ =
side opposite θ =
3 π π
4
hypotenuse 5 4. [−1,1] ;
−
2 ,
2
8. Let θ = sin −1
− 1
, then sin θ
1
= − .
Because sin θ
5. [−1,1] ; [0,π ]
2
2
is negative, θ is in quadrant IV. 6. (−∞, ∞) ;
− π
, π
2 2
7. − π
, π
2 2
8. [0,π ]
Use the Pythagorean Theorem to find x.
9. − π
, π
2 2
x2 + (−1)2 = 22
x2 + 1 = 4
x2 = 3
10. false
x = 3
Use values for x and r to find the exact value.
cos sin −1
− 1
= cos θ = x
= 3
2
r 2
Full file at https://testbank123.eu/Solutions-Manual-for-Trigonometry-2nd-Edition-Blitzer
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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
= −
is
3
,
= −
= −
Exercise Set 2.3
7. Let θ = cos−1 3
, then cos θ = 3
. The only angle
1. Let θ = sin −1 1 , then sin θ =
1 . The only angle in
2 2
2 2 in the interval [0, π ] that satisfies cos θ = 3 is
π .
the interval − π
, π
that satisfies sin θ = 1
is π
. 2 6
2 2 2 6
π − 3 π
Thus,
θ = π
, or sin −1 1
= π
. Thus θ = , or cos
1 = . 6 2 6
6 2 6
8. Let θ = cos−1 2
, then cos θ = 2
. The only angle2. Let θ = sin
−1 0 , then sin θ = 0 . The only angle in the 2 2
interval − π
, π
that satisfies sin θ = 0 is 0. Thus
in the interval [0, π ] that satisfies cosθ = 2
is π
. 2 2 2 4
θ = 0 , or sin −1
0 = 0 .
3. Let θ = sin −1 2
, then sin θ = 2
. The only angle
Thus θ = π
, or cos−1 2
= π
. 4 2 4
2 2 9. Let θ = cos−1 −
2 , then cosθ = −
2 . The only
in the interval − π
, π
that satisfies sin θ = 2
is 2 2
2 2
2 angle in the interval [0, π ] that satisfies
π . Thus
θ = π
, or sin −1 2
= π
.
cosθ = − 2
is 3π
. Thus θ = 3π
, or4 4 2 4 2 4 4
cos−1 − 2
= 3π
.4. Let θ = sin
−1 3 , then sin θ =
3 . The only angle in
2 42 2
the interval − π
, π
that satisfies sin θ = 3
is
−1
3 3 2 2
2 10. Let θ = cos
−
2
, then cosθ = −
2 .
π . Thus θ =
π , or sin−1 3
= π
.
The only angle in the interval [0, π ] that3 3 2 3
satisfies cosθ = − 3
is 5π
. Thus θ = 5π
, or
5. Let θ = sin −1
− 1
, then sin θ 1
. The only angle 2 6 6
2
2
cos−1 −
3 =
5π .
in the interval − π
, π
that satisfies sin θ 1 2 6
2 2 2
− π
. Thus θ
π , or sin −1 −
1 = −
π .
11. Let θ = cos−1
0 , then cosθ = 0 . The only angle in= −
6 6 2 6 the interval [0, π ] that satisfies cosθ = 0 is
π .
2
6. Let θ = sin −1 − 3
, then sin θ = − . Thus θ = π
, or cos−1 0 = π
. 2 2 2 2
The only angle in the interval − π
, π
that satisfies −1 2 2
12. Let θ = cos 1 , then cosθ = 1 . The only angle in the
3 sin θ = −
2
is − π
. Thus θ π
3 3
interval [0, π ] that satisfies cosθ = 1 is 0 .
Thus θ = 0 , or cos−1
1 = 0 .
or sin−1 − 3
= − π
.
2 3
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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
Scientific Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
sin −1
0.3
Radian
0.30 0.3 SIN−1
= −
= −
13. Let θ = tan −1 3
, then tan θ = 3
. The only angle in the interval − π
, π
that satisfies tan θ = 3
is π
. Thus3 3
2 2
3 6
θ = π
, or tan−1 3 = π
.
6 3 6
14. Let θ = tan −1
1 , then tanθ = 1 . The only angle in the interval − π
, π
that satisfies tanθ = 1 is π
.
2 2
4
Thus θ = π
, or tan−1 1 = π
.
4 4
15. Let θ = tan −1
0 , then tan θ = 0 . The only angle in the interval − π
, π
that satisfies tan θ = 0 is 0. Thus θ = 0 , or
2 2
tan −1
0 = 0 .
16. Let θ = tan−1
(−1) , then tan θ = −1 . The only angle in the interval − π
, π
that satisfies tan θ = −1 is − π
. Thus
2 2
4
θ π
, or tan−1 (−1) = − π
.4
17. Let θ = tan−1 (−
4
3 ) , then tan θ = − 3 . The only angle in the interval − π
, π
that satisfies tan θ = − 3 is
− π
. 2 2
3
Thus θ
π , or tan
−1 (−
3 ) = − π
.3 3
18. Let θ = tan−1
− 3
, then tan θ = − 3
. The only angle in the interval − π
, π
that satisfies tan θ = − 3
is − π
. 3 3 2 2 3 6
Thus θ π
, or tan−1 − 3
= − π
.= −
6 3 6
19.
Graphing Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
sin −1
0.3
Radian
0.30 SIN−1 0.3 ENTER
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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
Scientific Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
sin −1
0.47
Radian 0.47 SIN−1
0.49
Scientific Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
sin −1
(−0.32)
Radian
–0.33 SIN−1 0.32 + −
Scientific Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
sin−1
(−0.625)
Radian 0.625 + SIN−1
–0.68
Scientific Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
cos−1 3
Radian
−1
1.19
Graphing Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
cos−1 3
Radian
−1
1.19
20.
Graphing Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
sin −1
0.47
Radian SIN−1
0.47 ENTER
0.49
21.
Graphing Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
sin −1
(−0.32)
Radian
–0.33 SIN−1 − 0.32 ENTER
22.
−
Graphing Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
sin−1
(−0.625)
Radian SIN−1 − 0.625 ENTER
–0.68
23.
8
3 ÷ 8 = COS
8
COS ( 3 ÷ 8 ) ENTER
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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
130 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 130
Scientific Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
cos−1 4
Radian
−1
1.11
Graphing Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
cos−1 4
Radian
−1
1.11
Scientific Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
cos−1 5
7
Radian
5 ÷ 7 = COS−1
1.25
Scientific Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
tan −1
(−20)
Radian
–1.52 TAN−1 20 + −
24.
9
4 ÷ 9 = COS
9
COS ( 4 ÷ 9 ) ENTER
25.
Graphing Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
cos−1 5
7
Radian
COS−1
( 5 ÷ 7 ) ENTER
1.25
26. Scientific Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
cos−1 7
10
Radian
7 ÷ 10 = COS−1
1.30
Graphing Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
cos−1 7
10
Radian
COS−1
( 7 ÷ 10 ) ENTER
1.30
27.
Graphing Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
tan −1
(−20)
Radian
–1.52 TAN−1 − 20 ENTER
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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
131 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 131
Scientific Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
tan −1
(−30)
Radian 30 + TAN−1
–1.54
Scientific Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
tan −1 (− 473 )
Radian
–1.52 473 + −
TAN−1
Scientific Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
tan−1 (− 5061) Radian
−1
–1.56
Graphing Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
tan−1 (− 5061) Radian
−1
–1.56
28.
−
Graphing Calculator Solution
Function
Mode
Keystrokes Display
(rounded to two places)
tan −1
(−30)
Radian TAN−1 − 30 ENTER
–1.54
29.
Graphing Calculator Solution
Function Mode Keystrokes Display
(rounded to two places)
tan −1 (− 473 )
Radian TAN−1
( −
473 ) ENTER
–1.52
30.
5061 + − TAN
TAN ( − 5061 ) ENTER
31. sin (sin−1 0.9) x = 0.9, x is in [−1, 1], so sin(sin −1 0.9) = 0.9
32.
33.
cos(cos−1
0.57)
x = 0.57, x is in [−1, 1],
so cos(cos−1 0.57) = 0.57
sin −1
sin π
3
x = π
, x is in − π
, π
, so sin −1
sin π
= π
3 2 2
3
3
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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
132 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 132
2 2
3
4
34.
cos−1
cos 2π
41.
tan −1 tan 2π
3
3
x = 2π
, x is in [0, π ],
2π
π π 3
so cos−1
cos 2π
= 2π
x = , x is not in − , , x is in the domain of 3
3
3
−1
2π −1 π
35.
sin −1
sin 5π
tan x, so tan tan
= tan
(− 3 ) = − 3
6
−1
3π
x = 5π
, x is not in − π
, π
, x is in the domain of
42. tan tan
6 2 2
x = 3π
, x is not in − π
, π
,
4 2 2 sin x, so sin
−1 sin
5π = sin
−1 1 =
π
x is in the domain of tan x 6
2
6
so tan −1
tan 3π
= tan −1
(−1) = − π
4
4
36.
cos−1
cos 4π
3
x = 4π
, x is not in [0, π ], 3
43. sin −1
(sin π )
π π x is in the domain of cos x, x = π , x is not in − , ,
so cos−1
cos 4π
= cos−1
− 1
= 2π 2 2
3
2
3
x is in the domain of sin x, so
37.
tan ( tan−1
125) x = 125, x is a real number, so tan ( tan−1 125) = 125
44.
sin−1
(sin π ) = sin−1
0 = 0
cos−1
(cos 2π ) x = 2π , x is not in [0, π ],
x is in the domain of cos x,−1 −1
38.
39.
tan(tan−1
380)
x = 380, x is a real number,
so tan(tan−1 380) = 380
tan −1 tan − π
45.
so cos (cos 2π ) = cos 1 = 0
sin (sin−1 π )
x = π , x is not in [−1, 1] , so sin (sin−1 π ) is not
6 defined.
x π
, x is in − π
, π
, so −1= −
46.
cos(cos 3π )6 2 2 x = 3π , x is not in [−1, 1]
tan −1 tan − π
= − π
so cos(cos−1
3π ) is not defined.
6
6
40.
tan −1
tan − π
3
x π
, x is in − π
, π
,
= −
3 2 2
so tan −1
tan − π
= − π
3
3
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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
133 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 133
= −
5
47. Let θ = sin −1 4
, then sin θ = 4
. Because sin θ is 50. Let θ = sin −1 5 then sin θ =
5 .
5 5 13 13
positive, θ is in the first quadrant.
x2 + y2 = r 2
x2 + 42 = 52
x2 = 25 −16 = 9 x = 3
cos sin
−1 4 = cosθ =
x =
3
because sin θ is positive, θ is in the first quadrant.
x2 + y2 = r 2
x2 + 52 = 13
2
x2 = 144
x = 12
cot sin
−1 5 = cot θ =
x =
12
5
r 5 13 y 5
48. Let θ = tan−1 7
, then tan θ = 7
.
51. Let θ = sin −1
− 3
, then sin θ
3
. Because sin θ24 24
5
5
Because tan θ is positive, θ is in the first quadrant.
r 2 = x2 + y2
r 2 = 72 + 242
r 2 = 625 r = 25
is negative, θ is in quadrant IV.
x2 + y2 = r
2
x2 + (−3)
2 = 52
x2 = 16 x = 4
sin tan
−1 7 = sin θ =
y =
7 −1
3
y 3 24
r 25
tan sin
− = tan θ = = −
5 x 4
49. Let θ = cos−1 5 , then cosθ =
5
. Because cosθ is
−1 4 413 13 52. Let θ = sin − , then sin θ
= − .
5positive, θ is in the first quadrant.
x2 + y 2 = r 2
52 + y 2 = 132
y 2 = 169 − 25
y 2 = 144 y = 12
Because sin θ is negative, θ is in quadrant IV.
x2 + y2 = r 2
x2 + (−4)2 = 52
x2 = 9 x = 3
cos sin −1
− 4
= cosθ = x
= 3
tan cos−1 5
= tan θ = y
= 12 5 r 5
13
x 5
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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
134 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 134
.
= −
= −
4
53. Let, θ = cos−1 2 , then cosθ =
2 . Because cosθ
56. Let θ = sin
−1 −
1 , then sin θ
1
2 2
2
2
is positive, θ is in the first quadrant.
x2 + y2 = r 2 2 2 2
Because sin θ is negative, θ is in quadrant IV.
x2 + y2 = r 2
2 2 2( 2 ) + y = 2 x + (−1) = 2
y2 = 2
y = 2
x2 = 3
x = 3
sec sin −1 −
1 = secθ =
r =
2 =
2 3
sin cos−1 2
= sin θ = y
= 2
2 x 3 3
2 r 2
54. Let θ = sin −1 1
, then sin θ = 1
.
57. Let θ = cos−1
− 1
, then cosθ
1 . Because
2 2
3
3
Because sin θ is positive, θ is in the first quadrant.
x2 + y 2 = r 2
x2 + 12 = 22
x2 = 3
x = 3
cosθ is negative, θ is in quadrant II.
x2 + y2 = r
2
(−1)2 + y2 = 32
y2 = 8
y = 8
cos sin
−1 1 = cosθ =
x =
3 y = 2 2
2
r 2
Use the right triangle to find the exact value.
tan cos−1 − 1
= tan θ = y
= 2 2
= −2 255. Let θ = sin
−1 − 1
, then sin θ 1
. Because sin θ
3
x −1 4
= −
is negative, θ is in quadrant IV.
x2 + y
2 = r 2
x2 + (−1)
2 = 42
x2 = 15 x = 15
sec sin −1
− 1
= secθ = r
= 4
= 4 15
4
x 15 15
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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
135 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 135
. 2
−1
.
.
−1
2
= −
= −
= −
58. Let θ = cos−1
− 1
, then cosθ 1 x
2 + y
2 = r
2
4
4
x2 + (−
2 ) = 22
Because cosθ is negative, θ is in quadrant II. x
2 = 2
x = 2
2 r 2
sec sin − = secθ = = = 2
2 x 2
61. Let θ = tan −1
− 2
, then tan θ 2
3
3
x2
+ y2
= r 2
(−1)2
+ y2
= 42
y2
= 15
Because tan θ is negative, θ is in quadrant IV.
y = 15 θ
tan cos−1
− 1
= tan θ = y
= 15
= − 15 4
x −1
r 13
59. Let θ = cos−1 −
3 , then cosθ = −
3 . Because
r 2 = x2 + y2
r 2 = 32 + (−2)2
2 2 r 2 = 9 + 4cosθ is negative, θ is in quadrant II. r 2 = 13
r = 13
cos tan−1
− 2
= cosθ = x
= 3
= 3 13
3
r 13 13
62. Let θ = tan −1
− 3
, then tan θ 3
4
4
x2 + y2 = r
2
2 2 2
Because tan θ is negative, θ is in quadrant IV.
(− 3 ) + y = 2
y2 = 1 y = 1
3 r 2
csc cos − = cscθ = = = 2
2 y 1
r 2 = x2 + y2
60. Let θ = sin−1 − 2
, then sin θ = − .
r 2 = 42 + ( −3)2
2 2 r
2 = 16 + 9Because sin θ is negative, θ is in quadrant IV.
r 2 = 25 r = 5
sin tan
−1 −
3 = sin θ =
y =
−3 = −
3
4
r 5 5
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136 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 136
( ) 1 − x
−1
63. Let θ = cos−1 x , then cosθ = x = x
. 1
65. Let θ = sin −1
2 x , then sin θ = 2 x
y = 2x, r = 1
Use the Pythagorean Theorem to find x.
x2
+ (2x)2
= 12
x2 = 1 − 4 x
2
x = 1 − 4 x2
Use the Pythagorean Theorem to find the third
side, b.
x2 + b2 = 12
cos(sin −1
2 x) =
66. Let θ = cos−1
2 x.
1 − 4 x2
b2 = 1 − x2
Use the Pythagorean Theorem to find the third side, b.
b = 1 − x2
Use the right triangle to write the algebraic expression.
2
tan cos−1 x = tan θ = x
64. Let θ = tan −1
x , then tan θ = x = x
.
2 2 2
1 (2 x) + b = 1
b2 = 1 − 4 x2
b = 1 − 4 x2
−1 1 − 4 x 2
2sin (cos 2 x ) = 1
= 1 − 4 x
Use the Pythagorean Theorem to find the third side, c. 67. Let θ = sin −1 1
, then sin θ = 1
.
c2
= x2
+ 12 x x
c = x2 + 1
Use the right triangle to write the algebraic expression.
sin(tan −1 ) = sin θ
= x
x2 + 1
x x2 + 1
= ⋅
x2 + 1
x x2 + 1
= x2 + 1
x2 + 1 Use the Pythagorean Theorem to find the third side,
a.
a2 +12 = x2
2 2a = x
a =
−1
x2 −1
Use the right triangle to write the algebraic
expression.
1 x 2
− 1
cos sin = cosθ =
x
x
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137 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 137
−1
68. Let θ = cos−1 1 , then cosθ =
1 .
−1 x2 − 9 x2 − 9
x x 72. Let θ = sin , then sin θ = . x x
Use the Pythagorean Theorem to find the third
side, b.
12 + b2 = x2
Use the Pythagorean Theorem to find the third side,
a.
b2 = x2 − 1
a2 +
2
x2 − 9
= x2
b = x2 − 1
a2 = x2 − x2 + 9 = 9
Use the right triangle to write the algebraic expression.
sec cos−1 1
= secθ = x
= x
a = 3
Use the right triangle to write the algebraic expression.
x
1
2 −
cot sin −1 x 9
= 3
x x2 − 9
69.
cot tan
−1 x =
3
3
x
3
x2 − 9 3 x
2 − 9 = ⋅ =
70.
cot tan
−1 x =
2 x
2 − 9 x
2 − 9 x2 − 9
2
x
73. a.
y = sec x is the reciprocal of
y = cos x . The x-
71. Let θ = sin −1 x
, then sin θ = x
.
values for the key points in the interval [0, π ]
are 0, π
, π
, 3π
, and π . The key points arex2 + 4 x2 + 4 4 2 4
π 2 π 3π 2
(0, 1), , , , 0 , , − , and
4 2 2 4 2
(π , − 1) , Draw a vertical asymptote at x = π
. 2
Use the Pythagorean Theorem to find the third side,
Now draw our graph from (0, 1) through
π , 2
to ∞ on the left side of the
a. 4
a2 + x2 =
2
x2 + 4
asymptote. From −∞ on the right side of the
asymptote through
3π , − 2
to (π , − 1) .
a2 = x2 + 4 − x2 = 4 a = 2
Use the right triangle to write the algebraic
expression.
4
x
sec sin
secθ x
2 + 4
= =
x2
+ 4 2
b. With this restricted domain, no horizontal lineintersects the graph of y = sec x more than
once, so the function is one-to-one and has an
inverse function.
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138 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 138
2 2 ,
c. Reflecting the graph of the restricted secant
function about the line y = x, we get the graph 75.
of y = sec−1 x .
74. a. Two consecutive asymptotes occur at x = 0 and
76.
domain: [−1, 1];
range: [0, π]
x = π . Midway between x = 0 and x = π is
x = 0 and x = π . An x-intercept for the graph
is π
, 0 . The graph goes through the points
2
3π π , 1
and
, − 1
. Now graph the function
4
4
through these points and using the asymptotes. domain: [−1, 1];
range: π 3π
b. With this restricted domain no horizontal line
77.
intersects the graph of y = cot x more than
once, so the function is one-to-one and has an
inverse function. Reflecting the graph of the
restricted cotangent function about the line y =
x, we get the graph of
c.
y = cot −1
x .
78.
domain: [−2, 0];
range: [0, π]
domain: [−2, 0];
range:
− π
, π
2 2
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139 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 139
79.
80.
domain: (−∞, ∞);
range: (−π, π)
domain: (−∞, ∞);
83. 84.
domain: [−2, 2];
range: [0, π]
domain: [−2, 2];
range:
− 3π
, 3π
π π 2 2
range: − , 2 2
81.
82.
domain: (1, 3];
range: [−π, 0]
domain: [1, 3];
85. The inner function, sin −1
x, accepts values on the
interval [−1,1] . Since the inner and outer functions
are inverses of each other, the domain and range are
as follows.
domain: [−1,1] ; range: [−1,1]
86. The inner function, cos−1
x, accepts values on the
interval [−1,1] . Since the inner and outer functions
are inverses of each other, the domain and range are
as follows.
domain: [−1,1] ; range: [−1,1] 87. The inner function, cos x, accepts values on the
interval (−∞, ∞ ) . The outer function returns values
on the interval [0, π ] domain: (−∞, ∞ ) ; range: [0, π ]
88. The inner function, sin x, accepts values on the
interval (−∞, ∞ ) . The outer function returns values
range:
− π
, π
π π 2 2
on the interval − ,
2 2
domain: (−∞, ∞ ) ; range:
− π
, π
2 2
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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
140 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 140
x θ
5 tan −1 33
− tan −1 8
≈ 0.408 radians 5 5
10 tan −1 33
− tan −1 8
≈ 0.602 radians 10 10
15 tan −1 33
− tan −1 8
≈ 0.654 radians 15 15
20 tan −1 33
− tan −1 8
≈ 0.645 radians 20 20
25 tan −1 33
− tan −1 8
≈ 0.613 radians 25 25
2
89. The inner function, cos x, accepts values on the
interval (−∞, ∞ ) . The outer function returns values
95.
θ = 2 tan −1 21.634
28
≈ 1.3157 radians;
1.3157 180
≈ 75.4°on the interval −
π , π
π 2 2
domain: (−∞, ∞ ) ; range:
− π
, π
−1 21.634 2 2
96. θ = 2 tan 300
≈ 0.1440 radians;
0.1440 180
≈ 8.2°90. The inner function, sin x, accepts values on the π
interval (−∞, ∞ ) . The outer function returns values
on the interval [0, π ]
domain: (−∞, ∞ ) ; range: [0, π ]
97.
tan −1
b − tan −1
a = tan −1
2 − tan −1
0
≈ 1.1071 square units
−1 −1
−1 −1
91. The functions sin−1
x and cos−1
x
accept values on 98. tan b − tan a = tan 1 − tan (−2)
≈ 1.8925 square unitsthe interval [−1,1] . The sum of these values is always
π .
99. – 109. Answers may vary.
2 110.
y = sin −1 x
domain: [−1,1] ; range: π
y = sin −1
x + 2
92. The functions sin−1 x and cos−1 x accept values on
the interval [−1,1] . The difference of these values
range from − π 2
to 3π 2
domain: [−1,1] ; range:
− π
, 3π
2 2
93.
θ = tan −1 33
− tan −1 8
x x
The graph of the second equation is the graph of the first equation shifted up 2 units.
−1111. The domain of y = cos x is the interval
[–1, 1], and the range is the interval [0, π ] . Because
the second equation is the first equation with 1
subtracted from the variable, we will move our x
π π max to π , and graph in a − , π , by [0, 4, 1]
viewing rectangle.
2 4
94. The viewing angle increases rapidly up to about 16 feet, then it decreases less rapidly; about 16 feet;
when x = 15, θ = 0.6542 radians; when x = 17,
θ = 0.6553 radians.
The graph of the second equation is the graph of the first equation shifted right 1 unit.
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141 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 141
−1
112. y = tan −1
x
y = −2 tan −1
x
115.
The graph of the second equation is the graph of the
It seems sin−1
x + cos−1
x = π 2
for −1 ≤ x ≤ 1 .
first equation reversed and stretched. 116. does not make sense; Explanations will vary. Sample explanation: The cosine’s inverse is not a
113. The domain of y = sin−1 x is the interval function over that interval.
[–1, 1], and the range is − π
, π
. Because the 2 2
second equation is the first equation plus 1, and with 2 added to the variable, we will move our y max to 3, and move our x min to −π , and graph in a
−π , π
, π
by
117. does not make sense; Explanations will vary.
Sample explanation: Though this restriction works
for tangent, it is not selected simply because it is
easier to remember. Rather the restrictions are based
on which intervals will have inverses.
118. makes sense 2 2
[–2, 3, 1] viewing rectangle. 119. does not make sense; Explanations will vary.
Sample explanation:
sin −1 sin 5π
= sin −1 − 2
= − π
4 2 4
114.
The graph of the second equation is the graph of the first equation shifted left 2 units and up 1 unit.
y = tan −1
x
120.
121.
y = 2sin−1( x − 5) y
= sin ( x − 5) 2
sin y
= x − 5 2
x = sin y
+ 5 2
2sin −1
x = π
Observations may vary.
4
sin −1
x = π
8
x = sin π 8
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Section 2.3 Inverse Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
142 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 142
x y = 3cos 2π x coordinates
0 y = 3 cos(2π ⋅ 0) = 3cos 0 = 3 ⋅1 = 3
(0, 3)
1
4 y = 3cos
2π ⋅
1
π
= 3cos 2
= 3 ⋅ 0 = 0
1 , 0
4
1
2
1 y = 3cos 2π ⋅
2 = 3cos π = 3 ⋅ (−1) = −3
1 2
, − 3
3
4 y = 3cos
2π ⋅
3
3π 4
= 3cos 2
= 3 ⋅ 0 = 0
3 , 0
4
1 y = 3cos(2π ⋅1) = 3cos 2π
= 3 ⋅1 = 3
(1, 3)
x
= =
2
122. Prove: If x > 0, tan −1
x + tan −1 1
= π 126. The equation y = 3cos 2π x is of the form y = A cos Bx
x 2 with A = 3 and B = 2π . Thus, the amplitude is
Since x > 0, there is an angle θ with 0 < θ < π
as
A = 3 = 3 . The period is 2π
= 2π
= 1 . The quarter-2
shown in the figure. B 2π
1period is . The cycle begins at x = 0 . Add quarter-
4periods to generate x-values for the key points.
x = 0
x = 0 + 1
= 1
tan θ = x and tan π
− θ =
1 thus 1
4 4 2
x
x = + 1
= 1
tan −1
x = θ and tan −1 1
= π
− θ so 4 4 2
1 1 3 x = + = 2 4 4
tan −1
x + tan −1 1
= θ + π
− θ = π
x = 3
+ 1
= 1x 2 2 4 4
Evaluate the function at each value of x.123. Let α equal the acute angle in the smaller right
triangle.
tan α = 8 x
so tan −1 8 = α
x
tan(α + θ ) = 33
x
so tan −1 33 = α + θ
x
θ = α + θ − α = tan −1 33 − tan −1 8
x x
124. Because the tangent is negative and the cosine is
negative, θ lies in quadrant II. In quadrant II, x is negative and y is positive. Thus,
4
tan θ 2 y 2 = −
3 x −3 x = −3, y = 2
Furthermore,
r = x2 + y2 = (−3)2 + 22 = 9 + 4 = 13
Now that we know x, y, and r, we can find
sin θ and secθ .
sin θ = y
= 2
= 2
⋅ 13
= 2 13
Connect the five points with a smooth curve and
graph one complete cycle of the given function.
r 13 13 13 13
secθ = r
= 13
= − 13
x −3 3
125. 210° lies in quadrant III. The reference angle is
θ ′ = 210° − 180° = 30° .
sin 30° = 1 2
Because the sine is negative in quadrant III,
sin 210° = − sin 30° = − 1
. 2
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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
25
127. tan A = a
b
tan 22.3 =
a
12.1 a = 12.1 tan 22.3
a ≈ 4.96
cos A = b
c
cos 22.3 =
12.1
c
c = 12.1 cos 22.3
c ≈ 13.08
2. Using a right triangle, we have a known angle, an
unknown opposite side, a, and a known adjacent side.
Therefore, use the tangent function.
tan 85.4° = a 80
a = 80 tan 85.4° ≈ 994 The Eiffel tower is approximately 994 feet high.
3. Using a right triangle, we have an unknown angle, A,
a known opposite side, and a known hypotenuse.
Therefore, use the sine function.
sin A = 6.7
13.8
A = sin −1 6.7
≈ 29.0° 13.8
128.
tan θ = opposite
adjacent
tan θ = 18
25
θ = tan −1 18
θ ≈ 35.8
The wire makes an angle of approximately 29.0° with the ground.
4. Using two right triangles, a smaller right triangle
corresponding to the smaller angle of elevation drawn
inside a larger right triangle corresponding to the
larger angle of elevation, we have a known angle, an
unknown opposite side, a in the smaller triangle, b in
the larger triangle, and a known adjacent side in each
129. 10 cos π
x triangle. Therefore, use the tangent function. 6
a
amplitude: 10 = 10
tan 32° = 800
period: 2π
= 2π ⋅ 6
= 12
a = 800 tan 32° ≈ 499.9
tan 35° = b
π π 6
Section 2.4
Check Point Exercises
1. We begin by finding the measure of angle B. Because
C = 90° and the sum of a triangle’s angles is 180°, we
see that A + B = 90°. Thus, B = 90° – A = 90° – 62.7° = 27.3°. Now we find b. Because we have a known angle, a
known opposite side, and an unknown adjacent side,
use the tangent function.
tan 62.7° = 8.4 b
b = 8.4
≈ 4.34 tan 62.7°
Finally, we need to find c. Because we have a known
angle, a known opposite side and an unknown
hypotenuse, use the sine function.
sin 62.7° = 8.4 c
c = 8.4
≈ 9.45 sin 62.7
In summary, B = 27.3°, b ≈ 4.34, and c ≈ 9.45.
800 b = 800 tan 35° ≈ 560.2
The height of the sculpture of Lincoln’s face is 560.2 – 499.9, or approximately 60.3 feet.
5. a. We need the acute angle between ray OD and the north-south line through O. The measurement of this angle is given to be 25°. The angle is measured from the south side
of the north-south line and lies east of the north-
south line. Thus, the bearing from O to D is
S 25°E.
b. We need the acute angle between ray OC and
the north-south line through O. This angle measures 90° − 75° = 15°.
This angle is measured from the south side of
the north-south line and lies west of the north-
south line. Thus the bearing from O to C is S
15° W.
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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
4
° =
6. a. Your distance from the entrance to the trail
system is represented by the hypotenuse, c, of a
right triangle. Because we know the length of
the two sides of the right triangle, we find c
using the Pythagorean Theorem. We have
c2 = a2 + b2 = (2.3)
2 + (3.5)2 = 17.54
c = 17.54 ≈ 4.2
You are approximately 4.2 miles from the
entrance to the trail system.
8. We begin by identifying values for a and ω.
d = 12 cos π
t, a = 12 and ω = π
. 4 4
a. The maximum displacement from the rest
position is the amplitude. Because
a = 12, the maximum displacement is 12 centimeters.
b. The frequency, f, is
ω π
π 1 1 f = = = ⋅ =
b. To find your bearing from the entrance to the 2π 2π 4 2π 8
trail system, consider a north-south line passing
through the entrance. The acute angle from this The frequency is
1
8
cm per second.
line to the ray on which you lie is 31° + θ . Because we are measuring the angle from the
c. The time required for one cycle is the period.
south side of the line and you are west of the
period = 2π
= 2π
= 2π ⋅ 4
= 8
πentrance, your bearing from the entrance is ω 4
π
S (31° + θ ) W. To find θ , Use a right triangle
and the tangent function.
tan θ = 3.5 2.3
θ = tan −1 3.5
≈ 56.7° 2.3
Thus, 31° + θ = 31° + 56.7° = 87.7°. Your
bearing from the entrance to the trail system is S
87.7° W.
The time required for one cycle is 8 seconds.
Concept and Vocabulary Check 2.4
1. sides; angles 2. north; south
7. When the object is released (t = 0), the ball’s
distance, d, from its rest position is 6 inches down. Because it is down, d is negative: when t = 0, d = –6. Notice the greatest distance from
rest position occurs at t = 0. Thus, we will use the
3. simple harmonic; a ; 2π
; ω
Exercise Set 2.4
ω
2π
equation with the cosine function, y = a cos ωt, to 1. Find the measure of angle B. Because
model the ball’s motion. Recall that a is the
maximum distance. Because the ball initially moves
down, a = –6. The value of ω can be found using the formula for the period.
period = 2π
= 4 ω
2π = 4ω
ω = 2π
= π
4 2
Substitute these values into d = a cos wt. The equation for the ball’s simple harmonic motion is
d = −6 cos π
t. 2
C = 90°, A + B = 90°. Thus, B = 90° − A = 90° − 23.5° = 66.5°.
Because we have a known angle, a known adjacent
side, and an unknown opposite side, use the tangent
function.
tan 23.5° = a 10
a = 10 tan 23.5° ≈ 4.35 Because we have a known angle, a known adjacent
side, and an unknown hypotenuse, use the cosine
function.
cos 23.5 10 c
c = 10
≈ 10.90 cos 23.5°
In summary, B = 66.5°, a ≈ 4.35, and c ≈ 10.90.
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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
50.2
2. Find the measure of angle B. Because C = 90°,
A + B = 90°. Thus, B = 90° − A = 90° − 41.5° = 48.5°.
5. Find the measure of angle A. Because
C = 90°, A + B = 90°. Thus, A = 90° − B = 90° −16.8° = 73.2° .
Because we have a known angle, a known adjacent side, and an unknown opposite side, use the tangent
Because we have a known angle, a known opposite side and an unknown adjacent side, use the tangent function.
function.
tan 41.5° = a
tan 16.8° = 30.5
a
20 a = 20 tan 41.5° ≈ 17.69
Because we have a known angle, a known adjacent side, and an unknown hypotenuse, use the cosine function.
a = 30.5
≈ 101.02 tan 16.8°
Because we have a known angle, a known opposite
side, and an unknown hypotenuse, use the sine function.
30.5
cos 41.5° = 20
sin 16.8° = c
c
c = 20
≈ 26.70 cos 41.5°
In summary, B = 48.5°, a ≈ 17.69 , and c ≈ 26.70 .
3. Find the measure of angle B. Because
c = 30.5
≈ 105.52 sin16.8°
In summary, A = 73.2°, a ≈ 101.02, and c ≈ 105.52.
6. Find the measure of angle A. Because C = 90°,
A + B = 90°.
C = 90°, A + B = 90°. Thus, A = 90° − B = 90° − 23.8° = 66.2° .
Thus, B = 90° − A = 90° − 52.6° = 37.4° .
Because we have a known angle, a known
hypotenuse, and an unknown opposite side, use the
sine function.
sin 52.6 = a
54
Because we have a known angle, a known opposite side, and an unknown adjacent side, use the tangent function.
tan 23.8° = 40.5
a
a 40.5
≈ 91.83
a = 54 sin 52.6° ≈ 42.90
Because we have a known angle, a known
hypotenuse, and an unknown adjacent side, use the
cosine function.
cos 52.6° = b 54
b = 54 cos 52.6° ≈ 32.80 In summary, B = 37.4°, a ≈ 42.90, and b ≈ 32.80.
= tan 23.8°
Because we have a known angle, a known opposite
side, and an unknown hypotenuse, use the sine
function.
sin 23.8° = 40.5
c
c = 40.5
≈ 100.36 sin 23.8°
4. Find the measure of angle B. Because C = 90°, A + B = 90°.
In summary, A = 66.2°, a ≈ 91.83 , and c ≈ 100.36 .
Thus, B = 90° − A = 90° − 54.8° = 35.2° .
Because we have a known angle, a known hypotenuse,
and an unknown opposite side, use the sine function.
sin 54.8° = a 80
a = 80 sin 54.8° ≈ 65.37
Because we have a known angle, a known hypotenuse, and an unknown adjacent side, use the cosine function.
cos 54.8 = b 80
b = 80 cos 54.8° ≈ 46.11
In summary, B = 35.2°, a ≈ 65.37 , and c ≈ 46.11 .
7. Find the measure of angle A. Because we have a known hypotenuse, a known opposite side, and an unknown angle, use the sine function.
sin A = 30.4 50.2
A = sin −1 30.4
≈ 37.3°
Find the measure of angle B. Because C = 90°, A + B = 90°. Thus,
B = 90° − A ≈ 90° − 37.3° = 52.7° .
Use the Pythagorean Theorem.
a2 + b2 = c2
(30.4)2 + b2 = (50.2)2
b2 = (50.2)2 − (30.4)2 = 1595.88
b = 1595.88 ≈ 39.95
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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
In summary, A ≈ 37.3°, B ≈ 52.7°, and b ≈ 39.95.
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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
65.8
7
24.7
17.6
8. Find the measure of angle A. Because we have a
known hypotenuse, a known opposite side, and an
unknown angle, use the sine function.
sin A = 11.2 65.8
A = sin −1 11.2
≈ 9.8°
Find the measure of angle B. Because C = 90°,
A + B = 90°. Thus, B = 90° − A ≈ 90° − 9.8° = 80.2° .
Use the Pythagorean Theorem.
a2 + b2 = c2
(11.2)2 + b2 = (65.8)2
b2 = (65.8)
2 − (11.2)2 = 4204.2
b = 4204.2 ≈ 64.84
11. Find the measure of angle A. Because we have a
known hypotenuse, a known adjacent side, and
unknown angle, use the cosine function.
cos A = 2 7
A = cos−1 2
≈ 73.4°
Find the measure of angle B. Because
C = 90°, A + B = 90°. Thus, B = 90° − A ≈ 90° − 73.4° = 16.6° .
Use the Pythagorean Theorem.
a2 + b2 = c2
a2 + (2)2 = (7)2
a2 = (7)
2 − (2)2 = 45
a = 45 ≈ 6.71
In summary, A ≈ 9.8°, B ≈ 80.2°, and b ≈ 64.84.
9. Find the measure of angle A. Because we have a
known opposite side, a known adjacent side, and an
unknown angle, use the tangent function.
tan A = 10.8 24.7
A = tan −1 10.8
≈ 23.6°
Find the measure of angle B. Because
C = 90°, A + B = 90°. Thus, B = 90° − A ≈ 90° − 23.6° = 66.4° . Use the Pythagorean Theorem.
c2 = a2 + b2 = (10.8)
2 + (24.7)2 = 726.73
c = 726.73 ≈ 26.96
In summary, A ≈ 73.4°, B ≈ 16.6°, and a ≈ 6.71.
12. Find the measure of angle A. Because we have a
known hypotenuse, a known adjacent side, and an
unknown angle, use the cosine function.
cos A = 4 9
A = cos−1 4
≈ 63.6° 9
Find the measure of angle B. Because C = 90°,
A + B = 90°.
Thus, B = 90° − A ≈ 90° − 63.6° = 26.4° .
Use the Pythagorean Theorem.
a2 + b2 = c2
a2 + (4)2 = (9)2
In summary, A ≈ 23.6°, B ≈ 66.4°, and a2 = (9)
2 − (4)2 = 65
c ≈ 26.96.
10. Find the measure of angle A. Because we have a known opposite side, a known adjacent side, and an
a =
In summary,
65 ≈ 8.06
A ≈ 63.6°, B ≈ 26.4° , and a ≈ 8.06 .
unknown angle, use the tangent function.
tan A = 15.3 17.6
A = tan −1 15.3
≈ 41.0°
Find the measure of angle B. Because C = 90°, A + B = 90°.
Thus, B = 90° − A ≈ 90° − 41.0° = 49.0° .
Use the Pythagorean Theorem.
c2 = a2 + b2 = (15.3)
2 + (17.6)2 = 543.85
c = 543.85 ≈ 23.32
13. We need the acute angle between ray OA and the north-south line through O. This angle measure
90° − 75° = 15°. This angle is measured from the
north side of the north-south line and lies east of the
north-south line. Thus, the bearing from O and A is
N 15° E. 14. We need the acute angle between ray OB and the
north-south line through O. This angle measures
90° − 60° = 30° . This angle is measured from the
north side of the north-south line and lies west of the north-south line. Thus, the bearing from O to B is N
In summary, A ≈ 41.0°, B ≈ 49.0° , and c ≈ 23.32 . 30° W.
15. The measurement of this angle is given to be 80°.
The angle is measured from the south side of the
north-south line and lies west of the north-south line.
Thus, the bearing from O to C is S 80° W.
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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
16. We need the acute angle between ray OD and the 19. When t = 0, d = 0. Therefore, we will use the
north-south line through O. This angle measures equation with the sine function, y = a sin ωt, to
90° − 35° = 55° . This angle is measured from the
south side of the north-south line and lies east of the north-south line. Thus, the bearing from O to D is S 55° E.
17. When the object is released (t = 0), the object’s
distance, d, from its rest position is 6 centimeters
down. Because it is down, d is negative: When
model the object’s motion. Recall that a is the
maximum distance. Because the object initially
moves down, and has an amplitude of 3 inches,
a = –3. The value of ω can be found using the formula for the period.
period = 2π
= 1.5 ω
2π = 1.5ωt = 0, d = −6. Notice the greatest distance from rest
position occurs at t = 0. Thus, we will use the ω = 2π
= 4π
1.5 3equation with the cosine function, y = a cos ωt to Substitute these values into d = a sin ωt. Themodel the object’s motion. Recall that a is the
maximum distance. Because the object initially
moves down, a = –6. The value of ω can be found using the formula for the period.
equation for the object’s simple harmonic motion is
d = −3sin 4π
t. 3
20. When t = 0, d = 0. Therefore, we will use theperiod =
2π = 4
ω
equation with the sine function,
y = a sin ωt, to
2π = 4ω
ω = 2π
= π
4 2
Substitute these values into d = a cos ωt. The
equation for the object’s simple harmonic motion is
d = −6 cos π
t. 2
model the object’s motion. Recall that a is the
maximum distance. Because the object initially moves down, and has an amplitude of 5 centimeters,
a = –5. The value of ω can be found using the
formula for the period.
period = 2π
= 2.5 ω 2π = 2.5ω
18. When the object is released (t = 0), the object’s
distance, d, from its rest position is 8 inches down.
Because it is down, d, is negative: When
t = 0, d = –8. Notice the greatest distance from rest
position occurs at t = 0. Thus, we will use the
ω = 2π
= 4π
2.5 5
Substitute these values into d = a sin ωt . The
equation for the object’s simple harmonic motion is
4πequation with the cosine function, y = a cos ωt , to d = −5sin t .
5model the object’s motion. Recall that a is the
maximum distance. Because the object initially
moves down, a = –8. The value of ω can be found using the formula for the period.
period = 2π
= 2 ω 2π = 2ω
21. We begin by identifying values for a and ω.
d = 5 cos π
t, a = 5 and ω = π
2 2 a. The maximum displacement from the rest
position is the amplitude. Because a = 5, the
maximum displacement is 5 inches.
ω = 2π
= π ω
π 2
π 1 12
Substitute these values into d = a cos ωt .
The equation for the object’s simple harmonic motion
is d = −8 cos π t .
b. The frequency, f, is
The frequency is 1 4
f = = = ⋅ = . 2π 2π 2 2π 4
inch per second.
c. The time required for one cycle is the period.
period = 2π
= 2π
= 2π ⋅ 2
= 4
ω π π 2
The time required for one cycle is 4 seconds.
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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
22. We begin by identifying values for a and ω .
d = 10 cos 2π t, a = 10 and ω = 2π
a. The maximum displacement from the rest position is the amplitude. Because a = 10, the maximum displacement is 10 inches.
25. We begin by identifying values for a and ω.
d = 1
sin 2t, a = 1
and ω = 2 2 2
a. The maximum displacement from the rest
position is the amplitude.
1
b. The frequency, f, is f = ω
= 2π
= 1 . Because a = ,
2 the maximum displacement is
2π 2πThe frequency is 1 inch per second.
c. The time required for one cycle is the period.
period = 2π
= 2π
= 1
1 inch.
2
b. The frequency, f, is
ω 2π f = ω
= 2
= 1
≈ 0.32.The time required for one cycle is 1 second.
23. We begin by identifying values for a and ω.
d = −6 cos 2π t, a = −6 and ω = 2π
a. The maximum displacement from the rest
2π 2π π The frequency is approximately 0.32 cycle per
second.
c. The time required for one cycle is the period.
period = 2π
= 2π
= π ≈ 3.14position is the amplitude. ω 2Because a = –6, the maximum displacement is 6 inches.
b. The frequency, f, is
f = ω
= 2π
= 1.
The time required for one cycle is approximately 3.14 seconds.
26. We begin by identifying values for a and ω .
1 12π 2π d = sin 2t, a = and ω = 2The frequency is 1 inch per second.
c. The time required for one cycle is the period.
period = 2π
= 2π
= 1
3 3 a. The maximum displacement from the rest
position is the amplitude.
ω 2π 1The time required for one cycle is 1 second.
Because a = , the maximum displacement is 3
1
24. We begin by identifying values for a and ω .
d = −8cos π
t, a = −8 and ω = π
inch . 3
ω
2 1
2 2 b. The frequency, f, is f = = = ≈ 0.32 .2π 2π π
a. The maximum displacement from the rest
position is the amplitude.
Because a = –8, the maximum displacement is 8 inches.
The frequency is approximately 0.32 cycle per second.
c. The time required for one cycle is the period.
period = 2π
= 2π
= π ≈ 3.14
ω π ω 2
b. The frequency, f, is f = = 2 = 1
. The time required for one cycle is2π 2π 4
approximately 3.14 seconds.
The frequency is 1
4
inch per second.
c. The time required for one cycle is the period.
period = 2π =
2π = 2π ⋅
2 = 4
ω π π 2
The time required for one cycle is 4 seconds.
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Section 2.4 Applications of Trigonometric Functions Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
3
2
27. We begin by identifying values for a and ω.
33. x = 300
− 300
d = −5sin 2π
t, a = −5 and ω = 2π
3 3
tan 34° x ≈ 298
tan 64°
a. The maximum displacement from the rest position is the amplitude. Because a = –5, the maximum displacement is 5 inches.
b. The frequency, f, is
ω 2π
2π 1 1 f = = = ⋅ = .
34.
35.
x = 500
− 500
tan 20° tan 48° x ≈ 924
x = 400 tan 40° tan 20° tan 40° − tan 20°
x ≈ 257
2π 2π 3 2π 3
100 tan 43° tan 38°The frequency is
1
cycle per second. 36. x =
tan 43° − tan 38°3
c. The time required for one cycle is the period.
x ≈ 482
π period =
2π =
2π = 2π ⋅
3 = 3
37. d = 4 cos π t − 2 ω 2π 2π
3
The time required for one cycle is 3 seconds.
28. We begin by identifying values for a and ω .
d = −4sin 3π
t, a = −4 and ω = 3π
2 2
a. The maximum displacement from the rest
position is the amplitude.
Because a = –4, the maximum displacement is 4 inches.
b. The frequency, f, is
a. 4 in.
1
ω 3π
3π 1 3 f = = = ⋅ = .
b. in. per sec 2
2π 2π 2 2π 4
c. 2 secThe frequency is
3 cycle per second.
4
c. The time required for one cycle is the period.
d. 1 2
period = 2π
= 2π
= 2π ⋅ 2
= 4
ω 3π 3π 3
38.
d = 3cos π t +
π 2 2
29.
30.
31.
32.
The required time for one cycle is 4
seconds. 3
x = 500 tan 40° + 500 tan 25° x ≈ 653
x = 100 tan 20° + 100 tan 8° x ≈ 50
x = 600 tan 28° − 600 tan 25° x ≈ 39
x = 400 tan 40° − 400 tan 28° x ≈ 123
a. 3 in.
b. 1
in. per sec 2
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4 2
c. 2 sec 42. 30 yd ⋅
3 ft = 90 ft
1 yd
d. − 1
2
Using a right triangle, we have a known angle, an
unknown opposite side, a, and a known adjacent
side. Therefore, use the tangent function.
39. d = −2sin π t
+ π
tan 38.7° = a
90a = 90 tan 38.7° ≈ 72
The height of the building is approximately 72 feet.
a. 2 in.
b. 1
in. per sec 8
c. 8 sec
d. −2
43. Using a right triangle, we have a known angle, a
known opposite side, and an unknown adjacent side,
a. Therefore, use the tangent function.
tan 23.7° = 305
a
a = 305
≈ 695 tan 23.7°
The ship is approximately 695 feet from the statue’s
base.
44. Using a right triangle, we have a known angle, a
known opposite side, and an unknown adjacent side,
a. Therefore, use the tangent function.
tan 22.3° = 200
a
a = 200
≈ 488
40. d 1
sin π t
− π
= − 2 4 2
a. 1
in. 2
b. 1
in. per sec 8
c. 8 sec
d. 2
41. Using a right triangle, we have a known angle, an
unknown opposite side, a, and a known adjacent side.
Therefore, use tangent function.
tan 21.3° = a 5280
a = 5280 tan 21.3° ≈ 2059 The height of the tower is approximately 2059 feet.
tan 22.3° The ship is about 488 feet offshore.
45. The angle of depression from the helicopter to point P
is equal to the angle of elevation from point P to the
helicopter. Using a right triangle, we have a known
angle, a known opposite side, and an unknown
adjacent side, d. Therefore, use the tangent function.
tan 36° = 1000
d
d = 1000
≈ 1376 tan 36°
The island is approximately 1376 feet off the coast. 46. The angle of depression from the helicopter to the stolen
car is equal to the angle of elevation from the stolen car to the helicopter. Using a right triangle, we have a known angle, a known opposite side, and an unknown adjacent side, d. Therefore, use the tangent function.
tan 72° = 800 d
d = 800
≈ 260 tan 72°
The stolen car is approximately 260 feet from a point
directly below the helicopter.
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23
40
47. Using a right triangle, we have an unknown angle, A,
a known opposite side, and a known hypotenuse.
Therefore, use the sine function.
sin A = 6 23
A = sin −1 6
≈ 15.1°
The ramp makes an angle of approximately 15.1°
with the ground.
48. Using a right triangle, we have an unknown angle, A,
a known opposite side, and a known adjacent side.
Therefore, use the tangent function.
cos 53° = b
150 b = 150 cos 53° ≈ 90
The boat has traveled approximately 90 miles north
and 120 miles east.
52. Using a right triangle, we have a known angle, a
known hypotenuse, and unknown sides. To find the
opposite side, a, use the sine function.
sin 64° = a 40
a = 40 sin 64° ≈ 36 To find the adjacent side, b, use the cosine function.
btan A =
250
40
cos 64° = 40
A = tan −1 250
≈ 80.9°
The angle of elevation of the sun is approximately 80.9°.
49. Using the two right triangles, we have a known angle,
an unknown opposite side, a in the smaller triangle, b
in the larger triangle, and a known adjacent side in
each triangle. Therefore, use the tangent function.
tan 19.2° = a 125
a = 125 tan 19.2° ≈ 43.5
tan 31.7° = b
125
b = 125 tan 31.7° ≈ 77.2 The balloon rises approximately 77.2 – 43.5 or 33.7 feet.
50. Using two right triangles, a smaller right triangle
corresponding to the smaller angle of elevation drawn
inside a larger right triangle corresponding to the
larger angle of elevation, we have a known angle, an
unknown opposite side, a in the smaller triangle, b in
the larger triangle, and a known adjacent side in each
triangle. Therefore, use the tangent function.
tan 53° = a 330
a = 330 tan 53° = 437.9
tan 63° = b
330
b = 330 tan 63° ≈ 647.7 The height of the flagpole is approximately 647.7 – 437.9, or 209.8 feet (or 209.7 feet).
51. Using a right triangle, we have a known angle, a
known hypotenuse, and unknown sides. To find the
opposite side, a, use the sine function.
sin 53° = a 150
a = 150 sin 53° ≈ 120 To find the adjacent side, b, use the cosine function.
b = 40 cos 64° ≈ 17.5 The boat has traveled about 17.5 mi south and 36 mi east.
53. The bearing from the fire to the second ranger is N
28° E. Using a right triangle, we have a known angle, a known opposite side, and an unknown adjacent side, b. Therefore, use the tangent function.
tan 28° = 7 b
b = 7
≈ 13.2 tan 28°
The first ranger is 13.2 miles from the fire, to the nearest tenth of a mile.
54. The bearing from the lighthouse to the second ship is
N 34° E. Using a right triangle, we have a known
angle, a known opposite side, and an unknown
adjacent side, b. Therefore, use the tangent function.
tan 34° = 9 b
b = 9
≈ 13.3 tan 34°
The first ship is about 13.3 miles from the lighthouse,
to the nearest tenth of a mile.
55. Using a right triangle, we have a known adjacent
side, a known opposite side, and an unknown angle,
A. Therefore, use the tangent function.
tan A = 1.5 2
A = tan 1.5
≈ 37° 2
We need the acute angle between the ray that runs
from your house through your location, and the
north-south line through your house. This angle
measures approximately 90° − 37° = 53°. This angle
is measured from the north side of the north-south
line and lies west of the north-south line. Thus, the
bearing from your house to you is N 53° W.
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9
7
56. Using a right triangle, we have a known adjacent
side, a known opposite side, and an unknown angle,
59. The frequency, f, is f = ω
, so 2π
A. Therefore, use the tangent function. 1 ω
tan A = 6
9
A = tan −1 6
≈ 34°
We need the acute angle between the ray that runs
from the ship through the harbor, and the north-south
line through the ship. This angle measures 90° − 34° = 56° . This angle is measured from the
= 2 2π
ω = 1
⋅ 2π = π 2
Because the amplitude is 6 feet, a = 6. Thus, the equation for the object’s simple harmonic motion is
d = 6sin π t.
north side of the north-south line and lies west of the
north-south line. Thus, the bearing from the ship to
the harbor is N 56° W. The ship should use a bearing
of N 56° W to sail directly to the harbor.
57. To find the jet’s bearing from the control tower, consider
60. The frequency, f, is
1 =
ω 4 2π
ω = 1
⋅ 2π = π
4 2
f = ω
, so 2π
a north-south line passing through the tower. The acute angle from this line to the ray on which the jet lies is
35° + θ . Because we are measuring the angle from the
north side of the line and the jet is east of the tower, the
jet’s bearing from the tower is N (35° + θ ) E. To find θ ,
use a right triangle and the tangent function.
Because the amplitude is 8 feet, a = 8. Thus, the equation for the object’s simple harmonic motion is
d = 8sin π
t . 2
ω
tan θ = 7 5
θ = tan −1 7
≈ 54.5°
5
61. The frequency, f, is
264 = ω 2π
f = , so 2π
Thus, 35° + θ = 35° + 54.5° = 89.5°.
The jet’s bearing from the control tower is N 89.5° E.
58. To find the ship’s bearing from the port, consider a
ω = 264 ⋅ 2π = 528π Thus, the equation for the tuning fork’s simple
harmonic motion is d = sin 528π t.
north-south line passing through the port. The acute
angle from this line to the ray on which the ship lies
is 40° + θ . Because we are measuring the angle from
the south side of the line and the ship is west of the
62. The frequency, f, is
98,100, 000 = ω
2π
f = ω
, so 2π
port, the ship’s bearing from the port is
S (40° + θ ) W . To find θ , use a right triangle and
the tangent function.
tan θ = 11 7
θ = tan −1 11 ≈ 57.5°
ω = 98,100, 000 ⋅ 2π = 196, 200, 000π
Thus, the equation for the radio waves’ simple
harmonic motion is d = sin 196, 200, 000π t . 63. – 69. Answers may vary.
−0.1x
Thus, 40° + θ = 40° + 57.5 = 97.5° . Because this
angle is over 90° we subtract this angle from 180° to find the bearing from the north side of the north- south line. The bearing of the ship from the port is N 82.5° W.
70. y = 4e cos 2 x
3 complete oscillations occur.
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71. y = −6e−0.09 x
cos 2π x Using the transitive property we have
(75 + d ) tan 20° = d tan 40° 75 tan 20° + d tan 20° = d tan 40° d tan 40° − d tan 20° = 75 tan 20° d (tan 40° − tan 20°) = 75 tan 20°
d = 75 tan 20° tan 40° − tan 20°
10 complete oscillations occur.
72. makes sense
Thus, h = d tan 40° 75 tan 20°
= tan 40° ≈ 48 tan 40° − tan 20°
The height of the building is approximately 48 feet.
78. Answers may vary.
73. does not make sense; Explanations will vary.
Sample explanation: When using bearings, the angle
must be less than 90 .
79.
r = x2 + y2
r = (−3)2 + 42 =
9 + 16 =
25 = 5
74. does not make sense; Explanations will vary.
Sample explanation: When using bearings, north
Now that we know x, y, and r, we can find the six
trigonometric functions of θ .
sin θ = y
= 4
and south are listed before east and west. r 5
cosθ = x
= −3
= − 3
75. does not make sense; Explanations will vary. Sample explanation: Frequency and Period are inverses of each other. If the period is 10 seconds
r 5 5
tan θ = y
= 4
= − 4
x −3 3
cscθ = r
= 5
then the frequency is 1
10
= 0.1 oscillations per
y 4 r 5 5
second.
76. Using the right triangle, we have a known angle, an unknown opposite side, r, and an unknown hypotenuse, r + 112. Because both sides are in terms of the variable r,
secθ = = = − x −3 3
cot θ = x
= −3
= − 3
y 4 4
we can find r by using the sine function.
sin 76.6° = r
80. 13π
− 4π = 13π
− 12π
= π
3 3 3 3 lies in quadrant I.
r + 112
sin 76.6°(r + 112) = r Because the tangent is positive in quadrant I, π
r sin 76.6° + 112 sin 76.6° = r r − r sin 76.6° = 112 sin 76.6°
r (1 − sin 76.6°) = 112 sin 76.6°
tan = 3 . 3
o 46r =
112 sin 76.6° ≈ 4002 81. sin 26 =
c1 − sin 76.6° The Earth's radius is approximately 4002 miles.
c sin 26o = 46 46
77. Let d be the adjacent side to the 40° angle. Using the
right triangles, we have a known angle and unknown
sides in both triangles. Use the tangent function.
82.
c = ≈ 105 yd sin 26
o
sec x cot x = 1
⋅ cos x
= 1
or csc x
tan 20° = h 75 + d
cos x sin x sin x
h = (75 + d ) tan 20°
83. tan x csc x cos x = sin x
⋅ 1
⋅ cos x
= 1
Also, tan 40° = h d
cos x sin x 1
h = d tan 40°
84. sec x + tan x = 1
+ sin x
= 1 + sin x
cos x cos x cos x
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154 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 154
x coordinates
0 (0, –2)
π
4
π , 0 4
π
2
π , 2 2
3π
4
3π , 0 4
π (π , − 2)
x coordinates
0 (0, 0)
π
8
π , 3
8
π
4
π , 0
4
3π
8
3π , − 3
8
π
2
(2π , 0)
Chapter 2 Review Exercises 2. The equation y = −2 cos 2x is of the form
1. The equation
y = 3sin 4x is of the form
y = A sin Bx y = A cos Bx with A = –2 and B = 2. The amplitude
2π =
2π = π
Thewith A = 3 and B = 4. The amplitude is A = 3 = 3. is A = −2 = 2. The period is . B 2
The period is 2π
= 2π
= π
. The quarter-period is quarter-period is
π . The cycle begins at x = 0. Add
B 4 2 π
π 1 π 2 = ⋅ = . The cycle begins at x = 0. Add 4 2 4 8
quarter-periods to generate x-values for the key
points.
4 quarter-periods to generate x-values for the key points.
x = 0
π πx = 0
x = 0 + π
= π
8 8
x = π
+ π
= π
8 8 4
x = π
+ π
= 3π
4 8 8
x = 3π
+ π
= π
8 8 2
Evaluate the function at each value of x.
x = 0 + = 4 4
x = π
+ π
= π
4 4 2
x = π
+ π
= 3π
2 4 4
x = 3π
+ π
= π 4 4
Evaluate the function at each value of x.
Connect the five key points with a smooth curve
and graph one complete cycle of the given function.
Connect the five key points with a smooth curve
and graph one complete cycle of the given function.
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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
x
coordinates
0
(0, 2)
π
(π , 0)
2π
(2π , − 2)
3π
(3π , 0)
4π
(4π , 2)
x
coordinates
0
(0, 0)
3
2
3 , 1
2 2
3
(3, 0)
9
2
9 , −
1
6
(6, 0)
3. The equation y = 2 cos 1
x 2
is of the form
4. The equation y = 1
sin π
x is of the form 2 3
y = A cos Bx with A = 2 and B = 1
. The amplitude 2
y = A sin Bx with A = 1 2
and B = π
. The amplitude 3
2π 2π 1 1 2π =
2π =
π ⋅
3 =is A = 2 = 2. The period is
B =
1 2
= 2π ⋅ 2 = 4π . is A = = . The period is 2 2 B π
3
2 6. π
The quarter-period is 4π
= π . The cycle begins at 4
x = 0. Add quarter-periods to generate x-values for the key points. x = 0
x = 0 + π = π
x = π + π = 2π
x = 2π + π = 3π
x = 3π + π = 4π Evaluate the function at each value of x.
The quarter-period is 6
= 3
. The cycle begins at 4 2
x = 0. Add quarter-periods to generate x-values for the key points.
x = 0
x = 0 + 3
= 3
2 2
x = 3
+ 3
= 3 2 2
x = 3 + 3
= 9
2 2
x = 9
+ 3
= 6 2 2
Evaluate the function at each value of x.
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
2 2
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
x coordinates
0 (0, 0)
1
2
1 , − 1
2
1 (1, 0)
2
2
2 (2, 0)
x
coordinates
0
(0, 3)
3π
2
3π , 0 2
3π
(3π , −3)
9π
2
9π , 0
6π
(6π , 3)
5. The equation y = − sin π x is of the form
6. The equation y = 3cos x
is of the form
y = A cos Bxy = A sin Bx with A = –1 and B = π . The amplitude 3
is A = −1 = 1. The period is 2π
= 2π
= 2. The with A = 3 and B =
1 . The amplitude is
A = 3 = 3.B π 3
quarter-period is 2
= 1
. The cycle begins at x = 0.
The period is 2π =
2π = 2π ⋅ 3 = 6π . The quarter-
4 2 Add quarter-periods to generate x-values for the key points.
period is
B
6π =
3π
1 3
. The cycle begins at x = 0. Add
x = 0
x = 0 + 1
= 1
2 2
x = 1
+ 1
= 1 2 2
x = 1 + 1
= 3
2 2
x = 3
+ 1
= 2 2 2
4 2 quarter-periods to generate x-values for the key points.
x = 0
x = 0 + 3π
= 3π
2 2
x = 3π
+ 3π
= 3π 2 2
x = 3π + 3π
= 9π
2 2Evaluate the function at each value of x.
9π 3πx = + = 6π
2 2 Evaluate the function at each value of x.
3 3
, 1
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
2
Connect the five key points with a smooth curve and graph one complete cycle of the given function.
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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
x coordinates
π (π , 0)
3π
2
3π
2 , 2
2π (2π , 0)
5π
2
5π , − 2
2
3π (3π , 0)
x coordinates
−π (−π , − 3)
π −
2 − π
, 0
2
0 (0, 3)
π
2
π , 0
2
π (π , − 3)
= −
7. The equation y = 2sin( x − π ) is of the form 8. y = −3cos( x + π ) = −3cos( x − (−π ))
y = A sin(Bx − C) with A = 2, B = 1, and C = π . The The equation y = −3cos( x − (−π )) is of the form
amplitude is A = 2 = 2. The period is y = A cos(Bx − C ) with A = –3, B = 1, and C = −π .
2π =
2π = 2π . The phase shift is
C = π
= π . The The amplitude is A = −3 = 3.
B 1 B 1
The period is 2π =
2π = 2π . The phase shift is
quarter-period is 2π
= π
. B 1
4 2 C =
−π = −π
The quarter-period is
2π = π
. TheThe cycle begins at x = π . Add quarter-periods to B 1 .
4 2generate x-values for the key points. x = π
x = π + π
= 3π
2 2
x = 3π
+ π
= 2π 2 2
x = 2π + π
= 5π
2 2
x = 5π
+ π
= 3π 2 2
Evaluate the function at each value of x.
cycle begins at x = −π . Add quarter-periods to
generate x-values for the key points. x = −π
x = −π + π
= − π
2 2
x π
+ π
= 0 2 2
x = 0 + π
= π
2 2
x = π
+ π
= π 2 2
Evaluate the function at each value of x.
Connect the five key points with a smooth curve
and graph one complete cycle of the given function.
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
x
coordinates
− π 8
− π
, 3
π
8
π , 0
8
3π
8
3π , −
3
5π
8
5π , 0
8
7π
8
7π ,
3
8 2
2
= −
9. y = 3
cos 2 x +
π =
3 cos
2 x −
− π Connect the five key points with a smooth curve and
2
4
2
4 graph one complete cycle of the given function.
The equation
y = 3
cos 2 x −
− π
is of2
4
the form
y = A cos(Bx − C ) with A = 3
, 2
B = 2, and C = − π
. The amplitude is A = 3 =
3 .
4 2 2
The period is 2π
= 2π
= π . The phase shift is
B 2 −π
C = 4 = −
π ⋅
1 = − π
. The quarter-period is π
. 5 π 5 π
B 2 4 2 8 4 10. y = sin 2 x + = sin 2x − −
The cycle begins at x π
. Add quarter-periods to 2 2 2 2
= − 5
π 8 The equation y = sin 2x − −
is ofgenerate x-values for the key points. 2 2
π x = −
8
the form
y = A sin(Bx − C ) with
π
A = 5
, 2
5 5π π π x = − + =
8 4 8 B = 2, and C = − . The amplitude is A = = .
2 2 2
x = π
+ π
= 3π
8 4 8
The period is
π
2π =
2π B 2
= π . The phase shift is
C − π 1 π π
= = − ⋅ = − . The quarter-period is .x =
3π + π
= 5π
8 4 8 B 2 2 2 4 4
x = 5π
+ π
= 7π
8 4 8
The cycle begins at x π
. Add quarter-periods to 4
generate x-values for the key points.
Evaluate the function at each value of x. π x = −
8 2
4
π π x = − + = 0
4 4
π π
8 2
x = 0 + = 4 4
x = π
+ π
= π
4 4 2
x = π
+ π
= 3π
2 4 4 Evaluate the function at each value of x.
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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
x
coordinates
− π 4
(− π
, 0) 4
0 0,
5
2
π
4
π , 0
π
2
π , −
5 2 2
3π
4
3π , 0
4
x
coordinates
9
(9, 0)
21
2
21 , − 3 2
12
(12, 0)
27
2
27 , 3
2
15
(15, 0)
4
x = 9
x = 9 + 3
= 21
2 2
x = 21
+ 3
= 12 2 2
x = 12 + 3
= 27
2 2
x = 27
+ 3
= 15 2 2
Evaluate the function at each value of x.
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
11. The equation
y = −3sin π
x − 3π
is of
Connect the five key points with a smooth curve and
graph one complete cycle of the given function.
3
the form
y = A sin(Bx − C ) with A = –3,
B = π
, and C = 3π . The amplitude is 3
A = −3 = 3.
The period is 2π =
2π = 2π ⋅
3 = 6. The phase shift
B π π 3
is C
= 3π
= 3π ⋅ 3
= 9. The quarter-period is 6
= 3
.
B π π 4 2 3
The cycle begins at x = 9. Add quarter-periods to
generate x-values for the key points.
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160 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 160
x coordinates
0 (0, 1)
π
4
π , 2
4
π
2
π , 1
2
3π
4
3π , 0
π (π , 1)
x
coordinates
0
(0, 0)
3π
2
3π , − 2
2
3π
(3π , − 4)
9π
2
9π , − 2
6π
(6π , 0)
12. The graph of y = sin 2x + 1 is the graph of y = sin 2 x
13. The graph of y = 2 cos 1
x − 2 is the graph ofshifted one unit upward. The period for both
functions is 2π
= π . The quarter-period is π
. The
3
y = 2 cos 1
x shifted two units downward. The period2 4 3
cycle begins at x = 0. Add quarter-periods to generate x-values for the key points.
x = 0
for both functions is 2π 1 3
= 2π ⋅ 3 = 6π . The quarter-
x = 0 + π
= π
period is 6π
= 4
3π . The cycle begins at x = 0. Add
24 4
x = π
+ π
= π
4 4 2
x = π
+ π
= 3π
2 4 4
x = 3π
+ π
= π 4 4
quarter-periods to generate x-values for the key
points.
x = 0
x = 0 + 3π
= 3π
2 2
x = 3π
+ 3π
= 3π 2 2
Evaluate the function at each value of x.
x = 3π + 3π
= 9π
2 2
x = 9π
+ 3π
= 6π 2 2
Evaluate the function at each value of x.
4
By connecting the points with a smooth curve we obtain one period of the graph.
2
By connecting the points with a smooth curve we
obtain one period of the graph.
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161 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 161
12 12
14. Select several values of x over the interval.
x
0 π
4
π
2
3π
4 π
5π
4
3π
2
7π
4 2π
y1 = 3 sin x 0 2.1 3 2.1 0 −2.1 −3 −2.1 0
y2 = cos x 1 0.7 0 −0.7 −1 −0.7 0 0.7 1
y = 3 sin x + cos x 1 2.8 3 1.4 −1 −2.8 −3 −1.4 1
15. Select several values of x over the interval.
x
0 π
4
π
2
3π
4 π
5π
4
3π
2
7π
4 2π
y1 = sin x 0 0.7 1 0.7 0 −0.7 −1 −0.7 0
y = cos 1
x 2
2
1
0.9
0.7
0.4
0
−0.4
−0.7
−0.9
−1
y = sin x + cos 1
x 2
1
1.6
1.7
1.1
0
−1.1
−1.7
−1.6
−1
16. a. At midnight x = 0. Thus, y = 98.6 + 0.3 sin π
⋅ 0 − 11π
= 98.6 + 0.3 sin −
11π
12
≈ 98.6 + 0.3(−0.2588) ≈ 98.52
The body temperature is about 98.52°F.
b. period: 2π
= 2π
= 2π ⋅ 12
= 24 hoursB π π
12
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162 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 162
12 12
12 12
12
c. Solve the equation
π x −
11π = π
12 12 2
π x = π
+ 11π
= 6π
+ 11π
= 17π
12 2 12 12 12 12
x = 17π
⋅ 12
= 17
x = 11
x = 11 + 6 = 17
x = 17 + 6 = 23
x = 23 + 6 = 29
x = 29 + 6 = 3512 π
The body temperature is highest for x = 17.
y = 98.6 + 0.3 sin π
⋅17 − 11π
= 98.6 + 0.3 sin π
= 98.6 + 0.3 = 98.9 2
17 hours after midnight, which is 5 P.M., the body temperature is 98.9°F.
d. Solve the equation
π x −
11π =
3π 12 12 2
Evaluate the function at each value of x. The key points are (11, 98.6), (17, 98.9), (23, 98.6), (29, 98.3), (35, 98.6). Extend the pattern to the
left, and graph the function for 0 ≤ x ≤ 24.
17. Blue:
This is a sine wave with a period of 480.
π x =
3π +
11π =
18π +
11π =
29π 12 2 12 12 12 12
x = 29π
⋅ 12
= 29
Since the amplitude is 1,
B = 2π
= 2π
= π
period 480 240
A = 1.
12 π
The equation is y = sin π
x.The body temperature is lowest for x = 29.
y = 98.6 + 0.3 sin π
⋅ 29 − 11π
12 12
Red:
240
This is a sine wave with a period of 640.
= 98.6 + 0.3 sin 3π Since the amplitude is 1, A = 1.
2
2π
2π π
= 98.6 + 0.3(−1) = 98.3°
B = = = period 640 320
29 hours after midnight or 5 hours after
midnight, at 5 A.M., the body temperature is 98.3°F.
The equation is y = sin π
x. 320
18. Solve the equations
e. The graph of y = 98.6 + 0.3 sin π
x − 11π
is
π 2 x = −
2 and 2 x =
π 2
π πof the form y = D + A sin(Bx − C ) with A = 0.3, − x = 2 x = 2
B = π
, C = 11π
, and D = 98.6. The amplitude 2 2
12 12
x π
x π
is A =
0.3 = 0.3. The period from part (b) = − =
4 4 Thus, two consecutive asymptotes occur at
is 24. The quarter-period is 24
= 6. The phase 4
π x = −
and x = π
.
C 11π
11π 12
shift is = = ⋅ = 11. The cycle
4 4
− π + π
B π 12
12 π x-intercept = 4 4 = 0
= 0 2 2
begins at x = 11. Add quarter-periods to generate x-values for the key points.
An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the tangent is 4, the points on the graph midway between an x-intercept
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163 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 163
= −
− π
−2π
π
2 2
and the asymptotes have y-coordinates of –4 and 4. 20. Solve the equations
Use the two consecutive asymptotes. x π
and x + π = − π
and x + π = π
4 2 2
x = π
, to graph one full period of
y = 4 tan 2x
from π x = − x =
π − π
4 2 2
− π
to π
.
x 3π
x π
= − = − 4 4 2 2
Continue the pattern and extend the graph another full period to the right.
Thus, two consecutive asymptotes occur at
x 3π π
= − and x = − . 2 2
− 3π − π x-intercept = 2 2 = = −π
2 2An x-intercept is −π and the graph passes through
(−π , 0) . Because the coefficient of the tangent is 1,
the points on the graph midway between an x-
intercept and the asymptotes have y-coordinates of –1
and 1. Use the two consecutive asymptotes,
3π π19. Solve the equations x = − and x = − , to graph one full period of
π π x = −
and π x = π
2 2
3π π4 2 4 2 y = tan( x + π ) from − to − .
x π 4
x π 4
= − ⋅ = ⋅2 π
x = −2
2 π x = 2
Continue the pattern and extend the graph another full period to the right.
Thus, two consecutive asymptotes occur at
x = –2 and x = 2.
x-intercept = −2 + 2
= 0
= 0 2 2
An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the tangent is –2, the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of 2 and –2.
Use the two consecutive asymptotes, x = –2 and x =
2, to graph one full period of
y = −2 tan π
x from –2 21. Solve the equations
4 x − π
= − π
and x − π
= π
to 2. Continue the pattern and extend the graph 4 2 4 2
another full period to the right. x
π π x
π π= − + = +
2 4 2 4
x π
x 3π
= − = 4 4
Thus, two consecutive asymptotes occur at
x π 3π
= − and x = − . 4 4
− π − 3π π
x-intercept = 4 4 = 2 = 2 2 4
An x-intercept is π
and the graph passes through 4
π , 0 . Because the coefficient of the tangent is –1,
4
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164 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 164
= −
x
= −
the points on the graph midway between an x-
intercept and the asymptotes have y-coordinates of 1
23. Solve the equations
π πand –1. Use the two consecutive asymptotes,
x = 0 and 2
x = π 2
x π 3π
x = 0 x = π ⋅ 2
= − and x = , to graph one full period of 4 4 π
y = − tan x − π
from − π
to 3π
. Continue the x = 2
4
4 4
pattern and extend the graph another full period to the
right.
Thus, two consecutive asymptotes occur at x = 0 and x = 2.
x-intercept = 0 + 2
= 2
= 1 2 2
An x-intercept is 1 and the graph passes through (1, 0). Because the coefficient of the cotangent is
− 1
, the points on the graph midway between an x- 2
intercept and the asymptotes have y-coordinates of
− 1
and 1
. Use the two consecutive asymptotes, 2 2
22. Solve the equations x = 0 and x = 2, to graph one full period of
3x = 0 and 3x = π y 1
cot π
x from 0 to 2. Continue the pattern and 2 2
x = 0 x = π 3
extend the graph another full period to the right.
Thus, two consecutive asymptotes occur at
x = 0 and x = π
. 3
0 + π π π
x-intercept = 3 = 3 =
2 2 6
An x-intercept is π
and the graph passes through 6
π , 0 .
6
24. Solve the equations
Because the coefficient of the tangent is 2, the points x + π
= 0 and x + π
= πon the graph midway between an x-intercept and the 2 2
asymptotes have y-coordinates of 2 and –2. Use the x = 0 −
π x = π − π
two consecutive asymptotes, x = 0 and x = π
, to 2 2
3 x
π x
π
graph one full period of
y = 2 cot 3x
from 0 to π
. 3
= − = 2 2
Thus, two consecutive asymptotes occur at
Continue the pattern and extend the graph another full period to the right.
π π = − and x = .
2 2
− π + π 0
x-intercept = 2 2 = = 0 2 2
An x-intercept is 0 and the graph passes through (0, 0). Because the coefficient of the cotangent is 2, the points on the graph midway between an x-intercept and the asymptotes have y-coordinates of 2 and –2.
Use the two consecutive asymptotes, x π
and2
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165 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 165
, 0 , , − 3 ,
x = π
, to graph one full period of
y = 2 cot x + π 26. Graph the reciprocal sine function, y = −2sin π x .
2
2
The equation is of the form
y = A sin Bx with
from − π
to π
. Continue the pattern and extend the
A = –2 and B = π .
2 2 amplitude: A = −2 = 2
graph another full period to the right.
period: 2π
= 2π
= 2 B π
Use quarter-periods, 2
= 1
, to find 4 2
x-values for the five key points. Starting with
x = 0, the x-values are 0, 1
, 2
1, 3
, 2 . Evaluating the 2
function at each value of x, the key points are (0, 0),
1 , − 2
, (1, 0) ,
3 , 2
, (2, 0) . Use these key points
2
2
25. Graph the reciprocal cosine function,
y = 3cos 2π x . to graph y = −2sin π x from 0 to 2. Extend the graph
The equation is of the form
and B = 2π .
y = A cos Bx
with A = 3 one cycle to the right. Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-intercepts, and use them as
amplitude: A = 3 = 3 guides to graph y = −2 csc π x.
period: 2π
= 2π
= 1 B 2π
Use quarter-periods, 1
, to find x-values for the five 4
key points. Starting with x = 0, the x-values are 0, 1
, 4
1 ,
3 , 1 . Evaluating the function at each value of x,
2 4 the key points are (0, 3),
1 1 3 , 0 , (1, 3) .
27. Graph the reciprocal cosine function,
y = 3cos( x + π ) . The equation is of the form 4
2
4 y = A cos(Bx − C ) with A = 3, B = 1, and C = −π .
Use these key points to graph y = 3cos 2π x from 0 amplitude: A = 3 = 3
to 1. Extend the graph one cycle to the right. Use the 2π =
2π = π
graph to obtain the graph of the reciprocal function.
Draw vertical asymptotes through the x-intercepts,
period: B
2 1
C −πand use them as guides to graph y = 3sec 2π x. phase shift: = = −π
B 1
Use quarter-periods, 2π
= π
, to find 4 2
x-values for the five key points. Starting with
x = −π , the x-values are −π , − π
, 2
0, π
, π . 2
Evaluating the function at each value of x, the key
points are (−π , 3) , − π
, 0 , (0, − 3) ,
2
π , 0 , (π , 3) . Use these key points to graph
2
y = 3cos( x + π ) from −π
to π . Extend the graph
one cycle to the right. Use the graph to obtain the
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166 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 166
3
2 2
= −
= −
graph of the reciprocal function. Draw vertical
asymptotes through the x-intercepts, and use them as 29. Let θ = sin
−1 1 , then sin θ = 1 .
guides to graph
y = 3 sec( x + π ).
The only angle in the interval − π
, π
that satisfies 2 2
sin θ = 1 is π
. Thus θ = π
, or sin −1 1 = π
.
2 2 2
28. Graph the reciprocal sine function,
y = 5
sin( x − π ) .
30. Let θ = cos−1
1 , then cosθ = 1 .
The only angle in the interval [0, π ] that satisfies
cosθ = 1 is 0 . Thus θ = 0 , or cos−1
1 = 0 .
31. Let θ = tan −1
1 , then tan θ = 1 .
The only angle in the interval − π
, π
that satisfies2 2 2
The equation is of the form
y = A sin(Bx − C ) with
π π −1 πtan θ = 1 is
. Thus θ = , or tan 1 = .A =
5 ,
2 B = 1, and C = π . 4 4 4
amplitude: A = 5
= 5 32. Let θ = sin −1 −
3 , then sin θ = − .
2 2
period: 2π
= 2π
= 2π
2 2
π π B 1 The only angle in the interval − , that satisfies
phase shift: C
= π
= π
3 π πB 1 sin θ = − is − . Thus θ = − , or
Use quarter-periods, 2π
= π
, to find 2 3 3
4 2 sin −1 − 3
= − π
.x-values for the five key points. Starting with x = π ,
the x-values are π , 3π
, 2π , 5π
, 3π . Evaluating the
2 3
2 2 33. Let θ = cos−1
− 1
, then cosθ 1
function at each value of x, the key points
2
2 .
are (π , 0), 3π
, 5
, (2π , 0) , 5π
, − 5
,
(3π , 0).
The only angle in the interval [0, π ] that satisfies 2 2
2 2
Use these key points to graph
y = 5
sin( x − π ) from
cosθ 1
is 2
2π . Thus θ =
2π , or
3 3
2 cos−1 −
1 =
2π .
π to 3π . Extend the graph one cycle to the right.
2
3 Use the graph to obtain the graph of the reciprocal function. Draw vertical asymptotes through the x-
−1
3 3intercepts, and use them as guides to graph 34. Let θ = tan
− 3
, then tan θ = − 3
.
y = 5
csc( x − π ).
π π 2 The only angle in the interval − , that satisfies
tan θ = − 3 3
is − π
. 6
2 2
Thus θ π
, or tan −1 − 3
= − π
.= −
6 3 6
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167 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 167
cos sin −1 2 = cos
π =
2 .
2
−1
Thus csc tan −1 3 = csc
π = 2 .
= −
35. Let θ = sin −1 2
, then sin θ = 2
. The only angle 40. Let θ = tan −1 3
, then tan θ = 3
2 2 4 4
in the interval − π
, π
that satisfies sin θ = 2
is
Because tan θ is positive, θ is in the first quadrant.
2 2 2
π .
4
Thus,
2 4 2
2 2 2
36. Let θ = cos−1 0 , then cosθ = 0 . The only angle in
the interval [0, π ] that satisfies cosθ = 0 is π
. 2
r = x + y
r 2 = 42 + 32
2
Thus, sin (cos−1 0) = sin π
= 1 . r = 25
r = 5
37. Let θ = sin −1 −
1 , then sin θ
1 . The only
cos
tan
−1 3 = cosθ =
x =
4= −
2
2
4 r 5
angle in the interval − π
, π
that satisfies 3 3 2 2
41. Let θ = cos−1 , then cosθ = . 5 5
sin θ 1
2
is − π
. 6
Because cosθ is positive, θ is in the first quadrant.
Thus, tan sin −1 −
1 = tan
− π
= − 3
. 2
6
3
38. Let θ = cos−1 −
3 , then cosθ = −
3 . The only
2 2 2
2 2 x + y = r
angle in the interval [0, π ] that satisfies
32 + y2 = 52
cosθ = − 3 2
is 5π
. 6
y2 = 25 − 9 = 16
3 5π 3
Thus, tan cos − = tan = − .
y = 16 = 4
2 6 3 sin
cos−1 3
= sin θ = y
= 4
5
r 5
39. Let θ = tan −1 3 , then tan θ =
3 .
3 3
The only angle in the interval − π
, π
that satisfies
2 2
tan θ = 3 3
is π
. 6
3 6
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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
168 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 168
.
.
= −
= −
= −
42. Let θ = sin −1 −
3 , then sin θ
3 44. Let θ = tan
−1 − 1
, 5
5
3
Because sin θ is negative, θ is in
quadrant IV.
Because tan θ is negative, θ is in quadrant IV and
x = 3 and y = −1 .
r 2 = x2 + y2
r 2 = 32 + (−1)
2
r 2 = 10
x2 + (−3)
2 = 52
r = 10
1
4
y −1 10sin tan
− −
= sin θ = = = −
x2 + y2 = r 2 5 r 10 10
x2 = 25 − 9 = 16
45.
x = π
, x is in − π
, π
, so sin −1
sin π
= π
3 2 2
3
3
x = 16 = 4
tan sin −1 −
3 = tan θ =
y = −
3
46.
x = 2π
, x is not in − π
, π
. x is in the domain of 5
x 4
3 2 2
sin x , so
43. Let θ = cos−1
− 4
, then cosθ 4
−1 2π −1 3 π
5
5 sin
sin
3 = sin
2 =
3
Because cosθ is negative, θ is in
quadrant II.
47.
sin −1
cos 2π
= sin −1 −
1
3
2
Let θ = sin −1 −
1 , then sin θ
1 . The only
2
2
angle in the interval − π
, π
that satisfies
2 2
sin θ
1 is − π
. Thus, θ
π , or
x2 + y2 = r
2
= − = − 2 6 6
(−4)2 + y2 = 52 sin
−1 cos
2π = sin
−1 − 1
= − π
. 3
2
6
y2 = 25 − 16 = 9
y = 9 = 3
48. Let θ = tan −1 x , then tan θ =
x .
2 2 Use the right triangle to find the exact value.
tan cos−1 −
4 = tan θ = −
3
5
4
r 2 = x2 + 22
r 2 = x2 + y2
r = x2 + 4
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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
169 Copyright © 2018 Pearson Education, Inc. Copyright © 2018 Pearson Education, Inc. 169
7
3.6
Use the right triangle to write the algebraic
expression.
2 x2 + 4
We have a known angle, a known opposite side, and
an unknown hypotenuse. Use the sine function.
sin 37.4° = 6
cos tan −1 x
= cosθ = 2
= 2
x2
4
x2 + 4
c +
49. Let θ = sin −1 1 , then sin θ =
1 .
c =
6 ≈ 9.88
sin 37.4°
x x In summary, A = 52.6°, a ≈ 7.85 , and c ≈ 9.88 .
Use the Pythagorean theorem to find the third side, b.
12 + b2 = x2
b2 = x2 − 1
b = x2 − 1
52. Find the measure of angle A. We have a known
hypotenuse, a known opposite side, and an unknown
angle. Use the sine function.
sin A = 2 7
A = sin −1 2
≈ 16.6°
Find the measure of angle B. Because C = 90° ,
A + B = 90° . Thus, B = 90° − A ≈ 90° −16.6° = 73.4°
We have a known hypotenuse, a known opposite side,
and an unknown adjacent side. Use the Pythagorean
theorem.Use the right triangle to write the algebraic expression. a
2 + b2 = c2
−1 1 x x x2 − 1 2
2 + b2 = 72
sec sin x
= secθ = 2
= 2 x − 1 x − 1
b2 = 72 − 22 = 45
50. Find the measure of angle B. Because C = 90° ,
b = 45 ≈ 6.71A + B = 90° . Thus, B = 90° − A = 90º −22.3° = 67.7°
We have a known angle, a known hypotenuse, and an In summary, A ≈ 16.6°, B ≈ 73.4° , and b ≈ 6.71 .
unknown opposite side. Use the sine function.
sin 22.3° = a 10
53. Find the measure of angle A. We have a known
opposite side, a known adjacent side, and an
unknown angle. Use the tangent function.
1.4a = 10 sin 22.3° ≈ 3.79
We have a known angle, a known hypotenuse, and an
tan A =
3.6
unknown adjacent side. Use the cosine function.
cos 22.3° = b 10
b = 10 cos 22.3° ≈ 9.25
In summary, B = 67.7°, a ≈ 3.79 , and b ≈ 9.25 .
51. Find the measure of angle A. Because C = 90° ,
A = tan −1 1.4
≈ 21.3°
Find the measure of angle B. Because C = 90° ,
A + B = 90° . Thus, B = 90° − A ≈ 90° − 21.3° = 68.7°
We have a known opposite side, a known adjacent
side, and an unknown hypotenuse.
Use the Pythagorean theorem.
A + B = 90° . Thus, A = 90° − B = 90° − 37.4° = 52.6° c2 = a2 + b2 = (1.4)2 + (3.6)2 = 14.92We have a known angle, a known opposite side, and an unknown adjacent side. Use the tangent function.
c = 14.92 ≈ 3.86
tan 37.4° = 6 a
a = 6
≈ 7.85 tan 37.4°
In summary, A ≈ 21.3°, B ≈ 68.7° , and c ≈ 3.86 .
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Chapter 2 Review Exercises Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
170 170
54. Using a right triangle, we have a known angle, an
unknown opposite side, h, and a known adjacent side.
Therefore, use the tangent function.
tan 25.6° = h 80
h = 80 tan 25.6°
≈ 38.3
The building is about 38 feet high.
55. Using a right triangle, we have a known angle, an
unknown opposite side, h, and a known adjacent side.
Therefore, use the tangent function.
tan 40° = h 60
h = 60 tan 40° ≈ 50 yd
The second building is 50 yds taller than the first.
Total height = 40 + 50 = 90 yd .
56. Using two right triangles, a smaller right triangle
corresponding to the smaller angle of elevation drawn
inside a larger right triangle corresponding to the
larger angle of elevation, we have a known angle, a
known opposite side, and an unknown adjacent side,
d, in the smaller triangle. Therefore, use the tangent
function.
59. Using a right triangle, we have a known angle, a
known adjacent side, and an unknown opposite side,
d. Therefore, use the tangent function.
tan 64° = d 12
d = 12 tan 64° ≈ 24.6
The ship is about 24.6 miles from the lighthouse.
60.
a. Using the figure, B = 58° + 32° = 90° Thus, use the Pythagorean Theorem to find the
distance from city A to city C.
8502 + 960
2 = b2
b2 = 722, 500 + 921, 600
b2 = 1, 644,100
tan 68 125
b = 1, 644,100 ≈ 1282.2° = d
The distance from city A to city B is about
d = 125
≈ 50.5 tan 68°
We now have a known angle, a known adjacent side,
and an unknown opposite side, h, in the larger
triangle. Again, use the tangent function.
tan 71° = h 50.5
h = 50.5 tan 71° ≈ 146.7
The height of the antenna is 146.7 − 125 , or 21.7 ft,
to the nearest tenth of a foot.
57. We need the acute angle between ray OA and the
north-south line through O. This angle measures 90° − 55° = 35° . This angle measured from the north
61.
1282.2 miles.
b. Using the figure,
tan A = opposite
= 960
≈ 1.1294 adjacent 850
A ≈ tan −1
(1.1294) ≈ 48°
180° − 58° − 48° = 74°
The bearing from city A to city C is S74°E.
d = 20 cos π
t 4
a = 20 and ω = π 4
a. maximum displacement:side of the north-south line and lies east of the north- south line. Thus the bearing from O to A is N35°E.
a = 20 = 20 cm
ω π
π
58. We need the acute angle between ray OA and the b. f = = 4 = ⋅ 1
= 1
north-south line through O. This angle measures 90° − 55° = 35° . This angle measured from the south
side of the north-south line and lies west of the north-
2π 2π
frequency: 1 8
4 2π 8
cm per second
south line. Thus the bearing from O to A is S35°W.
c. period: 2π =
2π
ω π 4
= 2π ⋅ 4
= 8 π
The time required for one cycle is 8 seconds.
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Chapter 2 Test Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
x coordinates
0 (0, 0)
π
4
π , 3
π
2
π 2
, 0
3π
4
3π , − 3 4
π (π , 0)
= −
62. d = 1
sin 4t 2
Chapter 2 Test
1. The equation
y = 3sin 2x is of the form
y = A sin Bx
a = 1
and ω = 4 with A = 3 and B = 2. The amplitude is A = 3 = 3.2
2π 2π πThe period is = = π . The quarter-period is .
a. maximum displacement: B 2 4
a = 1 =
1 cm
The cycle begins at x = 0. Add quarter-periods to generate x-values for the key points.
2 2
b. f = ω
= 4
= 2
≈ 0.64
x = 0
x = 0 + π
= π
2π 2π π 4 4frequency: 0.64 cm per second
c. period: 2π
= 2π
= π
≈ 1.57 ω 4 2
x = π
+ π
= π
4 4 2
π π 3πThe time required for one cycle is about 1.57
seconds.
63. Because the distance of the object from the rest position at t = 0 is a maximum, use the form
x = + = 2 4 4
x = 3π
+ π
= π 4 4
Evaluate the function at each value of x.
d = a cos ωt . The period is 2π ω
so,
2 = 2π
ω
ω = 2π
= π 2
Because the amplitude is 30 inches,
4
a = 30 .
because the object starts below its rest position
a = −30 . the equation for the object’s simple
harmonic motion is d = −30 cos π t .
64. Because the distance of the object from the rest
position at t = 0 is 0, use the form d = a sin ωt . The
period is 2π
so ω
Connect the five key points with a smooth curve
and graph one complete cycle of the given function.
5 = 2π
ω
ω = 2π
5
Because the amplitude is 1 4
inch,
a = 1
. a is 4
negative since the object begins pulled down. The equation for the object’s simple harmonic motion is
d 1
sin 2π
t. 4 5
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Chapter 2 Test Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
x coordinates
π
2
π , − 2
π (π , 0)
3π
2
3π , 2
2
2π (2π , 0)
5π
2
5π , −2
= −
2 2
2 2
2. The equation
y = −2 cos x − π
is of the form 3. Solve the equations
2
x π x π = − and =y = A cos(Bx − C ) with A = –2, B = 1, and 2 2 2 2
π πC =
π . The amplitude is
2 A = −2 = 2.
x = − ⋅ 2 2
x = −π
x = ⋅ 2 2
x = πThe period is
2π =
2π = 2π . The phase shift is
Thus, two consecutive asymptotes occur atB 1
x = −π and x = π . π
C = 2 =
π . The quarter-period is
2π = π
. x-intercept = −π + π
= 0
= 0B 1 2 4 2 2 2
The cycle begins at x = π
. Add quarter-periods to 2
generate x-values for the key points.
x = π
An x-intercept is 0 and the graph passes through (0,
0). Because the coefficient of the tangent is 2, the
points on the graph midway between an x-intercept
and the asymptotes have y-coordinates of –2 and 2.
2
x = π
+ π
= π
Use the two consecutive asymptotes, x = −π x = π , to graph one
x
and
2 2
x = π + π
= 3π
2 2
x = 3π
+ π
= 2π 2 2
x = 2π + π
= 5π
2 2
Evaluate the function at each value of x.
full period of y = 2 tan 2
from −π to π .
4. Graph the reciprocal sine function, y 1
sin π x. 2
2 The equation is of the form y = A sin Bx with A =
− 1
and B = π . 2
amplitude:
A = − 1
= 1
2 2
period: 2π
= 2π
= 2 B π
2
2 1
Connect the five key points with a smooth curve and Use quarter-periods, = , to find x-values for the
4 2
graph one complete cycle of the given function. five key points. Starting with x = 0, the
x-values are 0, 1
, 1, 3
, 2. Evaluating the function at 2 2
each value of x, the key points are
(0, 3), 1
, − 1
, (1, 0) , 3
, 1
, (2, 0) .
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Chapter 2 Test Chapter 2 Graphs of the Trigonometric Functions; Inverse Trigonometric Functions
= −
= −
. = −
Use these key points to graph y 1
sin π x 2
from 0 to 2. Use the graph to obtain the graph of the reciprocal function.
Draw vertical asymptotes through the x-intercepts, and use them as guides to graph
5. Select several values of x over the interval.
y 1
csc π x. 2
x
0 π
4
π
2
3π
4 π
5π
4
3π
2
7π
4 2π
y = 1
sin x 1
2
0
0.4
0.5
0.4
0
−0.4
−0.5
−0.4
0
y2 = 2 cos x 2 1.4 0 −1.4 −2 −1.4 0 1.4 2
y = 1
sin x + 2 cos x 2
2
1.8
0.5
−1.1
−2
−1.8
−0.5
1.1
2
6. Let θ = cos−1
− 1
, then cosθ 1
2
2
Because cosθ is negative, θ is in quadrant II.
x2 + y2 = r
2
(−1)2 + y2 = 22
y2 = 4 − 1 = 3
y = 3
tan cos−1 −
1 = tan θ =
y =
3 = − 3
2
x −1
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