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   y x 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 √ = 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 –

The Unit Circle   y   x r = 1

Quadrantal Angle  

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

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

Evaluate the six trigonometric functions of . θ SOLUTION Use the Pythagorean theorem to find the value of r. (–8)2 + (15)2 √ r = 64 + 225 √ = = 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 –

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 √ = 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

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

Reference Angle Relationships  

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°.

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 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

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

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. 7. 7π 9 – 7π 9 The angle – is coterminal with . The terminal side lies in Quadrant III, so θ' = – π = 11π 2π

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

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

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 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.

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

 

9.3 Assignment day 2 P. 574, 16-30 all