9.3 Evaluate Trigonometric Functions of Any Angle

How can you evaluate trigonometric functions of any angle?

What must always be true about the value of r?

Can a reference angle ever have a negative measure?

General Definitions of Trigonometric Functions•

x

y

Sometimes called circular functions

Let (–4, 3) be a point on the terminal side of an angle θ in standard position. Evaluate the six trigonometric functions of θ.

SOLUTION

Use the Pythagorean theorem to find the value of r.

x2 + y2√r = (–4)2 + 32√= = 25√ = 5Using x = –4, y = 3, and r = 5, you can write the following:

sin θ =yr =

35 cos θ =

xr =

45–

tan θ =yx =

34

– csc θ =ry =

53

sec θ =rx =

54– cot θ =

xy =

43

–

The Unit Circle•

r = 1

x

y

Quadrantal Angle

Use the unit circle to evaluate the six trigonometric functions of = 270°.θ

SOLUTION

Draw the unit circle, then draw the angle θ = 270° in standard position. The terminal side of θ intersects the unit circle at (0, –1), so use x = 0 and y = –1 to evaluate the trigonometric functions.

cos θ =xr =

01 = 0 undefined

undefined cot θ =xy

=0

–1tan θ =yx =

–10

sec θ =rx =

10

sin θ =yr

1= 1

–= –1 csc θ =

ry =

11– = –1

= 0

Evaluate the six trigonometric functions of .θ1.

SOLUTION

Use the Pythagorean Theorem to find the value of r.

x2 + y2√r = 32 + (–3)2√= = 18√ = 3√ 2

sin θ =yr cos θ =

xr

tan θ =yx =

33– csc θ =

ry

sec θ =rx cot θ =

xy =

33–

Using x = 3, y = –3 , and r = 3√ 2, you can write the following:

=3

–3√ 2 = –

3√ 23

= –1 = 3√ 23

– = –√ 2

3√ 23= = √ 2 = –1

= –2

√ 2= 2

√ 2

SOLUTION

Use the Pythagorean theorem to find the value of r.

(–8)2 + (15)2√r = 64 + 225√= = 289√ = 17

Evaluate the six trigonometric functions of .θ

sin θ =yr = 15

17 cos θ =xr = 8

17–

tan θ =yx =

158– csc θ =

ry =

1715

sec θ =rx =

178– cot θ =

xy = 8

15–

Using x = –8, y = 15, and r = 17, you can write the following:

Evaluate the six trigonometric functions of .θ

SOLUTION

Use the Pythagorean theorem to find the value of r.

x2 + y2√r = (–5)2 + (–12)2√= = 25 + 144√ = 13Using x = –5, y = –12, and r = 13, you can write the following:

sin θ =yr cos θ =

xr =

513

–

tan θ =yx = 12

5 csc θ =ry

sec θ =rx cot θ =

xy = 5

12

= 1213

–

= 1312–

= 135–

4. Use the unit circle to evaluate the six trigonometric functions of θ = 180°.

Draw the unit circle, then draw the angle θ = 180° in standard position. The terminal side of θ intersects the unit circle at (–1, 0), so use x = –1 and y = 0 to evaluate the trigonometric functions.

SOLUTION

cos θ =xr = –1

cot θ =xy

tan θ =yx = 0

–1

sec θ =rx

= –11

= –1= –11 = –1

0 undefined

sin θ =yr

0= 1 = 0

csc θ =ry = –1

0 undefined

Reference Angle Relationships•

Find the reference angle θ' for (a) θ = 5π3

and (b) θ = – 130°.

SOLUTION

a. The terminal side of θ lies in Quadrant IV. So, θ' = 2π – . 5π

3π3

=

b. Note that θ is coterminal with 230°, whose terminal side lies in Quadrant III. So, θ' = 230° – 180° + 50°.

9.3 Assignment

Page 574, 4-15 all

9.3 Evaluate Trigonometric Functions of Any Angle

• How can you evaluate trigonometric functions of any angle?

• What must always be true about the value of r?

• Can a reference angle ever have a negative measure?

Evaluating Trigonometric Functions

Reference Angle Relationships

Evaluate (a) tan ( – 240°).

SOLUTION

tan (–240°) = – tan 60° = – √ 3

a.

The angle – 240° is coterminal with 120°. The reference angle is θ' = 180° – 120° = 60°. The tangent function is negative in Quadrant II, so you can write:

30º

60º

3ll2

l

Evaluate (b) csc .17π6

SOLUTIONb. The angle is coterminal

with . The reference angle is θ' = π – = .

The cosecant function is positive in Quadrant II, so you can write:

17π65π

6 5π6

π6

csc = csc = 217π6

π6

30º

60º

3ll2

l30

6

Sketch the angle. Then find its reference angle.

5. 210°

The terminal side of θ lies in Quadrant III, so θ' = 210° – 180° = 30°

Sketch the angle. Then find its reference angle.

6. – 260°

– 260° is coterminal with 100°, whose terminal side of θ lies in Quadrant II, so θ' = 180° – 100° = 80°

Sketch the angle. Then find its reference angle.

7π9The angle – is coterminal with . The

terminal side lies in Quadrant III, so θ' = – π =

11π9

11π9

2π9

7.7π9

–

Sketch the angle. Then find its reference angle.15π4

8.

The terminal side lies in Quadrant IV, so θ' = 2π – = 15π

4π4

9. Evaluate cos ( – 210°) without using a calculator.

– 210° is coterminal with 150°. The terminal side lies in Quadrant II, which means it will have a negative value.

So, cos (– 210°) = – 2√ 3

30º

150º 30º

60º

3ll2

l

Robotics

The “frogbot” is a robot designed for exploring rough terrain on other planets. It can jump at a 45° angle and with an initial speed of 16 feet per second. On Earth, the horizontal distance d (in feet) traveled by a projectile launched at an angle θ and with an initial speed v (in feet per second) is given by:

d = v2

32 sin 2θ

How far can the frogbot jump on Earth?

SOLUTION

d = v2

32 sin 2θ

d = 162

32 sin (2 45°)

= 8

Write model for horizontal distance.

Substitute 16 for v and 45° for θ.

Simplify.

The frogbot can jump a horizontal distance of 8 feet on Earth.

A rock climber is using a rock climbing treadmill that is 10.5 feet long. The climber begins by lying horizontally on the treadmill, which is then rotated about its midpoint by 110° so that the rock climber is climbing towards the top. If the midpoint of the treadmill is 6 feet above the ground, how high above the ground is the top of the treadmill?

Rock climbing

SOLUTION sin θ =yr

sin 110° =y

5.254.9 y

Use definition of sine.

Solve for y.The top of the treadmill is about 6 + 4.9 = 10.9 feet above the ground.

Substitute 110° for θ and = 5.25 for r.2

10.5

9.3 Assignment day 2

P. 574, 16-30 all