4.4 Trigonometric Functions of Any Angle€¦ · 1 4.4 Trigonometric Functions of Any Angle r-r-r r...

1

4.4 Trigonometric Functions of Any Angle

r

-r

-r

r

y

x

every point on the circle satisfies to x2+y2=r2

is any angle in standard position

any circle of radius r

P(x,y)

P(x,y) is any point onthe terminal side of

y

x

2


r

-r

-r

r

y

x

P(x,y)y

x

sinθ=yr

cosθ=xr

tanθ=yx

cot θ=xy

cscθ=ry

secθ=rx

3


r

-r

-r

r

y

x

P(x,y)y

x

sinθ=yr

cosθ=xr

tanθ=yx

cot θ=xy

cscθ=ry

secθ=rx

+ +

- -

- +

- +

- +

+ -

+ +

- -

- +

- +

4


sinθ=yr

cosθ=xr

tanθ=yx

cot θ=xy

+ +

- -

- +

- +

- +

+ -

Examples:1) point (-12,5) lies on the terminal side of standard angle . Find the exact values of sin , cos , tan and cot .

5


sinθ=yr

cosθ=xr

tanθ=yx

cot θ=xy

+ +

- -

- +

- +

- +

+ -

Examples:1) point (-12,5) lies on the terminal side of standard angle . Find the exact values of sin , cos , tan and cot .

Solution:(1) find the radius:r

r

y

x

(−12,5)

sinα=yr=

−1213

-125

52+(−12)2=r2

25+144=r2

r2=169 r=13(2)

cosα=xr=513

tanα=yx=−

512

cotα=−125

6


tanθ=yx

cot θ=xy

secθ=rx

- +

+ -

- +

- +

Examples:2) Find tan

3π2, sec π ,cot π

2

7


tanθ=yx

cot θ=xy

secθ=rx

- +

+ -

- +

- +

Examples:2) Find

Solution:(1) use the unit circle

(radius = 1)

(2)

(0,1)

1

y

x

tan3π2, sec π ,cot π

2

π2

3 π2

π(-1,0)

(0,-1)

tan3π2

=yx=

−10

=undefined

sec π=rx=1

−1=−1

cot π2=xy=01=0

8


sinθ=yr

cosθ=xr

tanθ=yx

cot θ=xy

cscθ=ry

secθ=rx

+ +

- -

- +

- +

- +

+ -

+ +

- -

- +

- +

Examples:3) Given , in quadrant II,

find the exact value of each of the remaining trigonometric functions of .

cosβ=−35

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sinθ=yr

cosθ=xr

tanθ=yx

cot θ=xy

cscθ=ry

secθ=rx

+ +

- -

- +

- +

- +

+ -

+ +

- -

- +

- +

Examples:3) Given , in quadrant II,


5

cosβ=−35

-3

Solution:(1) find the radius

(2)

52=(−3)2+ y2y

y2=16 y=4

sinβ=yr=45

tanβ=yx=−

43

cotβ=−34

cscβ=ry=54

secβ=rx=−

53

10


sinθ=yr

cosθ=xr

tanθ=yx

cot θ=xy

cscθ=ry

secθ=rx

+ +

- -

- +

- +

- +

+ -

+ +

- -

- +

- +

Examples:4)


tanθ=512

,cosθ<0

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sinθ=yr

cosθ=xr

tanθ=yx

cot θ=xy

cscθ=ry

secθ=rx

+ +

- -

- +

- +

- +

+ -

+ +

- -

- +

- +

Examples:4)


tanθ=512

Solution:(1) is in the III quadrant, point (-12,-5)(2) the radius is:

(3)

(−5)2+(−12)2=r2

y r2=169 r=13

sinθ=yr=

−513

cot θ=125

cscθ=ry=−

135

secθ=rx=−

1312

,cosθ<0

(−12 ,−5) cosθ=xr=

−1213

12


y

x

Reference Angles

If is a non-acute angle in standard position that lies in a quadrant, then its reference angle is a positive angle ’ formed by the terminal side of and x-axis.

y

x

y

x

’

’’

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Reference Angles

If is a non-acute angle in standard position that lies in a quadrant, then its reference angle is a positive angle ’ formed by the terminal side of and x-axis.

The values of trigonometric functions of a given angle are the same as the values of the trigonometric functions of the reference angle ’, except possibly for the sign.

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Examples: use reference angles and trigonometric functions of special values table to evaluate

1)

2)

3)

sin4π3

cot(−π4 )

cos135o- +

- +

+ +

- -

- +

+ -

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1) ’ = 180º – 135º = 45º,

2)

3)

sin4π3

cot(−π4 )

cos135o- +

- +

+ +

- -

- +

+ -

cos135o=−cos45o=−√22

θ '=4π3

−π=π3

, is in the III quadrant

sin4π3

=−sin π3=−

√32

, is in the II quadrant

, is in the IV quadrant + cot is odd f.

θ '=π4

cot(−π4 )=−cot(

π4 )=−1

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4)

5)

6)

sec240o

cos7π4

sin14π3

- +

- +

- +

- +

+ +

- -

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4)

5)

6)

sec240o

cos7π4

sin14π3

- +

- +

- +

- +

+ +

- -, is in the II quadrant

14 π3

=4 π+2π3

θ '=π2π3

=π3

sin14π3

=sin π3=

√32

, is in the III quadrant

’ = 240º – 180º = 60º, sec240o=−sec60o=−2

7 π4

=π+3π4

θ '=2π−7π4

=π4

, is in the IV quadrant

cos7π4

=cos π4=

√22

4.4 Trigonometric Functions of Any Angle€¦ · 1 4.4 Trigonometric Functions of Any Angle r-r-r r...

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