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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. r = √ x2 + y2 = √ (–4)2 + 32 = √ 25 = 5 Using x = –4, y = 3, and r = 5, you can write the following: 3 4 y x sin θ = cos θ = = 5 =– 5 r r 3 5 y r tan θ = csc θ = =– 4 = 3 x y 5 r x 4 – sec θ = cot θ = = =– 4 x y 3 The Unit Circle y r=1 x Quadrantal Angle Use the unit circle to evaluate the six trigonometric functions of θ = 270°. 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. sin θ = cos θ = tan θ = 1 –1 y r csc θ = = 1 = –1 = – 1 = –1 r y 1 0 r x sec θ = = 0 undefined = 1 =0 x r –1 0 y x = 0 undefined cot θ = = –1 = 0 x y Evaluate the six trigonometric functions of θ. 1. SOLUTION Use the Pythagorean Theorem to find the value of r. r = √ x2 + y2 = √ 32 + (–3)2 = √ 18 = 3√ 2 Using x = 3, y = –3 , and r = 3√ 2, you can write the following: 3 y x √2 – – – 3 = √2 = sin θ = cos θ = = = 2 r 3√ 2 r 3√ 2 2 3 y r – – 3√ 2 = –√ 2 tan θ = csc θ = = = = –1 3 3 x y r x 3 3√ 2 sec θ = cot θ = =– = = –1 = √ 2 x 3 y 3 Evaluate the six trigonometric functions of θ. SOLUTION Use the Pythagorean theorem to find the value of r. r = √ (–8)2 + (15)2 = √ 64 + 225 = √ 289 = 17 Using x = –8, y = 15, and r = 17, you can write the following: y x 15 8 sin θ = cos θ = = 17 = – 17 r r y r 15 17 – tan θ = csc θ = = = 15 8 x y r x 17 8 – sec θ = – cot θ = = = 8 x y 15 Evaluate the six trigonometric functions of θ. SOLUTION Use the Pythagorean theorem to find the value of r. r = √ x2 + y2 = √ (–5)2 + (–12)2 = √ 25 + 144 = 13 Using x = –5, y = –12, and r = 13, you can write the following: y 12 sin θ = = – 13 r y tan θ = = 12 x 5 r 13 – sec θ = = 5 x x 5 – cos θ = = 13 r r 13 – csc θ = = 12 y x 5 cot θ = = y 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. y r y x sin θ = tan θ = sec θ = r x 0 = 1 =0 0 = –1 –1 = 1 = –1 cos θ = x –1 = r 1 csc θ = r y –1 = 0 undefined cot θ = x y –1 = 0 undefined = –1 Reference Angle Relationships 5π Find the reference angle θ' for (a) θ = 3 and (b) θ = – 130°. SOLUTION a. The terminal side of θ lies in Quadrant IV. π 5π = So, θ' = 2π – . 3 3 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 The angle – 240° is coterminal a. 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 2l l 3 60º l 17π Evaluate (b) csc . 6 SOLUTION b. The angle 17π is coterminal 6 5π with . The reference 6 5π π angle is θ' = π – = . 6 6 The cosecant function is positive in Quadrant II, so you can write: 30º π =2 6 csc 17π = csc 6 6 2l l 3 30 60º l 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. 7π – 7. 9 11π 7π The angle – 9 is coterminal with 9 . The terminal side lies in Quadrant III, 2π 11π so θ' = –π= 9 9 Sketch the angle. Then find its reference angle. 8. 15π 4 The terminal side lies in Quadrant IV, π 15π so θ' = 2π – = 4 4 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°) = – √ 3 2 30º 150º 30º 2l l 3 60º 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: 2 v d = 32 sin 2θ How far can the frogbot jump on Earth? SOLUTION 2 v sin 2θ d= Write model for horizontal distance 32 2 16 d = 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? y sin θ = SOLUTION Use definition of sine. r y 10.5 sin 110° = Substitute 110° for θ and = 5.25 for r. 5.25 2 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