9.3 Evaluate Trigonometric Functions of Any Angle

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 Functions y 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. 4 5 5 3 4 3 y x – sinθ cosθ = = = = 3 4 3 4 5 5 r r y r – tanθ cscθ = = = = x y r x – secθ – cotθ = = = = x y r √ 25 x2 + y2 (–4)2 + 32 √ √ = = = SOLUTION Use the Pythagorean theorem to find the value of r. = 5 Using x = –4, y = 3, and r =5, you can write the following:

4. The Unit Circle y x r = 1

5. Quadrantal Angle

6. Use the unit circle to evaluate the six trigonometric functions of = 270°. 1 0 r x secθ θ cosθ = = = = 0 1 x r –1 0 y x tanθ cotθ = = = = 0 –1 x y 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. 1 1 – y r sinθ cscθ = = = = – 1 1 r y SOLUTION = –1 = –1 undefined = 0 undefined = 0

7. Evaluate the six trigonometric functions of . Using x = 3, y =–3 , and r =3√2, you can write the following: θ 1. 3 3 y x 3 3 sinθ cosθ = = – – = = 3 3 r r 3√2 3√2 y r – tanθ = = – cscθ = = x y r x secθ = 3√ 2 – cotθ = = = x y r √ 18 x2 + y2 32 + (–3)2 √ √ = = 3√2 = 3√2 = = –√2 = √2 – 3 = = 3 2 2 √ 2 √ 2 SOLUTION Use the Pythagorean Theorem to find the value of r. = –1 = –1

8. Evaluate the six trigonometric functions of . θ 15 8 – = = 17 17 y x sinθ cosθ = = r r 15 17 – = = y r 8 15 tanθ cscθ = = x y r x 17 8 secθ cotθ = = – – = = x y 8 15 r (–8)2 + (15)2 √ = √ 289 64 + 225 √ = = SOLUTION Use the Pythagorean theorem to find the value of r. = 17 Using x = –8, y = 15, and r =17, you can write the following:

9. Evaluate the six trigonometric functions of . θ y x 5 sinθ cosθ = = – = r r 13 y r tanθ cscθ = = 12 x y = 5 r x secθ cotθ = = x y 5 = 12 r x2 + y2 √ = √ 25 + 144 (–5)2 + (–12)2 √ = = 13 13 12 – – – = = = 13 12 5 SOLUTION Use the Pythagorean theorem to find the value of r. = 13 Using x = –5, y = –12, and r =13, you can write the following:

10. r x secθ cosθ = = x r y x 0 tanθ cotθ = = = x y –1 0 –1 –1 –1 –1 = = = = = 1 0 1 0 1 y r sinθ cscθ = = r y 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. = 0 = –1 undefined = –1 undefined

11. Reference Angle Relationships

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

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

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

19. Sketch the angle. Then find its reference angle. 5. 210° 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. – 7. 7π 11π 11π 2π 7π 9 9 9 9 9 The angle – is coterminal with . The terminal side lies in Quadrant III, so θ' = – π = Sketch the angle. Then find its reference angle.

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

23. √ 3 2 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°) = – 30º 150º 30º 60º

24. sin 2θ d = v2 32 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: How far can the frogbot jump on Earth?

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

26. 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? y Substitute 110° for θ and = 5.25 for r. sin 110° = 5.25 10.5 sin θ = 2 4.9 y y r Rock climbing SOLUTION Use definition of sine. Solve for y. The top of the treadmill is about 6 + 4.9 = 10.9 feet above the ground.

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