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.

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

5csc 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 – . 5π3

π3

=

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 .17π

6SOLUTION

b. 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