1. 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?

2. General Definitions of Trigonometric Functionsy x
Sometimes called circular functions

3. 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 √
= 5
Using x = –4, y = 3, and r = 5, you can write the following:
sin θ = y r = 3 5
cos θ = x r = 4 5 –
tan θ = y x = 3 4 –
csc θ = r y = 5 3
sec θ = r x = 5 4 –
cot θ = x y = 4 3 –

4. The Unit Circle y x
r = 1

5. Quadrantal Angle

6. 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.
Use the unit circle to evaluate the six trigonometric functions of = 270°. θ
SOLUTION
sin θ = y r 1 = –
csc θ = r y = 1 –
= –1
= –1
cos θ = x r = 1
sec θ = r x = 1
= 0
undefined
tan θ = y x = –1 = –1
cot θ = x y
undefined
= 0

7. Evaluate the six trigonometric functions of .θ 1.
SOLUTION
Use the Pythagorean Theorem to find the value of r.
x2 + y2 √ r =
32 + (–3)2 √ = = 18 √
= 3√ 2
Using x = 3, y = –3 , and r = 3√ 2, you can write the following:
sin θ = y r = 3 –
3√ 2 = – 2
√ 2
cos θ = x r = –
3√ 2 3 = 2
√ 2
tan θ = y x = 3 –
csc θ = r y =
3√ 2 3 –
= –1
= –√ 2
sec θ = r x
cot θ = x y = 3 –
3√ 2 3 =
= √ 2
= –1

8. Evaluate the six trigonometric functions of .θ
SOLUTION
Use the Pythagorean theorem to find the value of r.
(–8)2 + (15)2 √ r = √ = =
289 √
= 17
Using x = –8, y = 15, and r = 17, you can write the following:
sin θ = y r = 15 17
cos θ = x r = 8 17 –
tan θ = y x = 15 8 –
csc θ = r y = 17 15
sec θ = r x = 17 8 –
cot θ = x y = 8 15 –

9. 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 √ = = √
= 13
Using x = –5, y = –12, and r = 13, you can write the following:
sin θ = y r = 12 13 –
cos θ = x r = 5 13 –
tan θ = y x
csc θ = r y = 12 5 = 13 12 –
sec θ = r x = 13 5 –
cot θ = x y = 5 12

10. 4. Use the unit circle to evaluate the six trigonometric functions of θ = 180°.
SOLUTION
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.
sin θ = y r = 1
cos θ = x r = –1 1
= 0
= –1
tan θ = y x = –1
csc θ = r y = –1
undefined
sec θ = r x = –1 1
cot θ = x y = –1
= –1
undefined

11. Reference Angle Relationships

12. Find the reference angle θ' for (a) θ =5π 3
and (b) θ = – 130°.
SOLUTION a.
The terminal side of θ lies in Quadrant IV.
So, θ' = 2π – 5π 3 π = b.
Note that θ is coterminal with 230°, whose terminal side lies in Quadrant III. So, θ' = 230° – 180° + 50°.

13. 9.3 Assignment
Page 574, 4-15 all

14. 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?

15. Evaluating Trigonometric Functions

16. Reference Angle Relationships

17. Evaluate (a) tan ( – 240°). SOLUTION 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:
tan (–240°) = – tan 60° = – √ 3
30º
60º

18. The angle is coterminal with . The reference angle is θ' = π – = .
Evaluate (b) csc
17π 6
SOLUTION b.
The angle is coterminal
with The reference
angle is θ' = π – =
The cosecant function is positive in Quadrant II, so you can write:
17π 6 5π π
30º
csc = csc = 2
17π 6 π
60º

19. Sketch the angle. Then find its reference angle.°
The terminal side of θ lies in Quadrant III,
so θ' = 210° – 180° = 30°

20. 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°

21. Sketch the angle. Then find its reference angle.7. 7π 9 – 7π 9
The angle – is coterminal with The terminal side lies in Quadrant III,
so θ' = – π =
11π 2π

22. Sketch the angle. Then find its reference angle.
15π 4 8.
The terminal side lies in Quadrant IV,
so θ' = 2π – =
15π 4 π

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

24. 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?

25. SOLUTION
d = v2 32
sin 2θ
Write model for horizontal distance.
d =
162 32
sin (2 45°)
Substitute 16 for v and 45° for θ.
= 8
Simplify.
The frogbot can jump a horizontal distance of 8 feet on Earth.

26. Rock climbing
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?
sin θ = y r
SOLUTION
Use definition of sine.
sin 110° = y
5.25
Substitute 110° for θ and = 5.25 for r. 2
10.5 y
Solve for y.
The top of the treadmill is about = 10.9 feet above the ground.

28. 9.3 Assignment day 2
P. 574, all