Third Edition Hughes-Hallett,
Transcript of Third Edition Hughes-Hallett,
1
Powerpoint slides copied from or based upon:
Connally,
Hughes-Hallett,
Gleason, Et Al.
Copyright 2007 John Wiley & Sons, Inc.
Functions Modeling Change
A Preparation for Calculus
Third Edition
CHAPTER 6
TRIGONOMETRIC FUNCTIONS
SECTION 6.6
OTHER TRIGONOMETRIC FUNCTIONS
The Tangent Function
Another useful trigonometric function is called
the tangent.
Suppose P = (x, y) is the point on the unit
circle corresponding to the angle θ. We define
the tangent of θ, or tan θ, by
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tany
x
The Tangent Function
The graphical interpretation of tan θ is as a
slope.
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tany
x
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0tan
0
y y ym
x x x
6Page 279 Example #1
Find the slope of the line passing through the
origin at an angle of
(a) 30°
(b) π/4
7Page 279 Example #1
Find the slope of the line passing through the
origin at an angle of
(a) 30°
(b) π/4
(a) In degree mode, tan(30) = .5773502692
8Page 279 Example #1
Find the slope of the line passing through the
origin at an angle of
(a) 30°
(b) π/4
(a) In degree mode, tan(30) = .5773502692
Thus, the slope of the line is = .5773502692
9Page 279 Example #1
Find the slope of the line passing through the
origin at an angle of
(a) 30°
(b) π/4
(b) In radian mode, tan (π/4) = 1
10Page 279 Example #1
Find the slope of the line passing through the
origin at an angle of
(a) 30°
(b) π/4
(b) In radian mode, tan (π/4) = 1
An angle of π/4 describes a ray halfway
between the x- and y-axes, or the line y = x,
which has slope of 1. (π/4 = 45°)
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Graph of the Tangent Function
Table 6.10 contains values of the tangent
function.
θ 0 30 60 90 120 150 180
θ 0 π/6 π/3 π/2 2π/3 5π/6 π
tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0
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Graph of the Tangent Function
As Θ goes from 0 to π/2, the slope (tangent)
goes from 0 to +∞.
θ 0 30 60 90 120 150 180
θ 0 π/6 π/3 π/2 2π/3 5π/6 π
tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0
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Graph of the Tangent Function
At precisely π/2 (90 ), this line is vertical, and
its slope is undefined.
θ 0 30 60 90 120 150 180
θ 0 π/6 π/3 π/2 2π/3 5π/6 π
tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0
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Graph of the Tangent Function
Thus, tan(π/2) is undefined and the graph of
y = tan θ has a vertical asymptote at θ = π/2.
θ 0 30 60 90 120 150 180
θ 0 π/6 π/3 π/2 2π/3 5π/6 π
tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0
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Graph of the Tangent Function
For θ between π/2 and π, the slopes are
negative.
θ 0 30 60 90 120 150 180
θ 0 π/6 π/3 π/2 2π/3 5π/6 π
tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0
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Graph of the Tangent Function
The line is very steep near π/2, but becomes
less steep as θ approaches π, where it is
horizontal.
θ 0 30 60 90 120 150 180
θ 0 π/6 π/3 π/2 2π/3 5π/6 π
tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0
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Graph of the Tangent Function
Thus, tan θ becomes less negative as θ
increases in the second quadrant and reaches
0 at θ = π.
θ 0 30 60 90 120 150 180
θ 0 π/6 π/3 π/2 2π/3 5π/6 π
tan θ 0 0.6 1.7 Undefined −1.7 −0.6 0
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Let's graph the tangent function on our
calculator:
y1=tan(x)
Window Value
Xmin 0
Xmax π
Xscl π/2
Ymin -5
Ymax 5
Yscl 1
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y1=tan(x)
Trace...
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y1=tan(x)
x=0, y=0
x=1, y=1.5574077
x=1.5, y=14.10142
x=1.55, y=48.078482
x=1.57, y=1,255.76
x=1.5707963, y=37,320,535!!
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Let's change the window settings:
y1=tan(x)
Window Value
Xmin -2π
Xmax 2π
Xscl π/2
Ymin -5
Ymax 5
Yscl 1
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y1=tan(x)
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For values of θ between π and 2π, observe
that: tan(Θ+π) = tanΘ
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tan(Θ+π) = tanΘ because the angles θ and (θ + π)
determine the same line through the origin, and
hence the same slope.
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tan(Θ+π) = tanΘ
Thus, y = tan θ has period π.
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Notice the differences between the tangent function
and the sinusoidal functions. The tangent function
has a period of π, whereas the sine and cosine both
have periods of 2π.
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The tangent function has vertical asymptotes at odd
multiples of π/2 (1π/2, 3π/2, 5π/2, etc.); the sine and
cosine have no asymptotes.
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Relationships Between the Trigonometric
Functions
There is a relationship among the three
trigonometric functions we have defined.
If (x, y) are the coordinates of the point P
determined by the angle θ on the unit circle,
we defined: cosθ = x and sinθ = y
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Since tan θ = y/x, if x ≠ 0, we have:
sintan
cos
The Unit Circle
The unit circle is the circle of radius one that is
centered at the origin.
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X
y
The Unit Circle
Since the distance from the point P with coordinates
(x, y) to the origin is 1, we have:
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X
y
2 2 1x y
The Unit Circle
Therefore:
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X
y
2 2 1x y
The Unit Circle
Therefore:
Substituting x = cos θ and y = sin θ gives:
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2 2 1x y
2 2cos sin 1
The Reciprocals of the Trigonometric
Functions: Secant, Cosecant, Cotangent
The reciprocals of the trigonometric functions
are given special names. Where the
denominators are not equal to zero, we have
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1secant =sec
cos
1cosecant = csc
sin
coscotangent = cot
sin
Use a graph of g(θ) = cos θ to explain the
shape of the graph of f(θ) = sec θ.
38Page 283 Example #5
Let's use our calculator:
Window: Zoom 7 (ZTrig)
39Page 283 Example #5
11 cos and 2
cosy y
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11 cos and 2
cosy y
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In the first quadrant, cos θ decreases from 1 to 0, so
the reciprocal of cos θ increases from 1 toward +∞.
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The values of cos θ are negative in the second
quadrant and decrease from 0 to −1 ...
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so the values of sec θ increase from −∞ to −1.
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y = cos θ is symmetric about the vertical line θ = π, so
the graph of f(θ) = sec θ is symmetric about the same
line.
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Thus, the graph of f(θ) = sec θ on the interval π≤θ≤2π
is the mirror image of the graph on 0≤θ≤π.
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Note that sec θ is undefined wherever cos θ = 0,
namely, at θ = π/2 and θ = 3π/2.
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The graph of f(θ) = sec θ has vertical asymptotes at
those values.
End of Section 6.6
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