Trigonometric Functions Chapter 4 Solution Key

Chapter 4: Trigonometric Functions 1. Find, if possible, the complement and supplement of 7π/10. C: none S: π - 7π/10 = 3π/10 2. Evaluate the six trigonometric functions of the angle. a. θ = 5π/2 sin(5π/2) = 1 cos(5π/2) = 0 tan(5π/2) = undefined csc(5π/2) = 1 sec(5π/2) = undefined cot(5π/2) = 0 b. θ = -2π/3 sin(-2π/3) = -√3/2 cos(-2π/3) = -1/2 tan(-2π/3) = √3 csc(-2π/3) = -2√3/3 sec(-2π/3) = -2 cot(-2π/3) = √3/3 3. Find the point (x, y) on the unit circle that corresponds to -17π/6. x = cos(-17π/6) = -√3/2 y = sin(-17π/6) = -1/2 (-√3/2, -1/2) 4. Evaluate the trigonometric function. a. sin(-π/6) = -1/2 b. csc(5π/2) = 1 c. tan(π/3) = √3 d. sec(4π) = 1 e. cot(π/2) = 0 f. cos(13π/4) = -√2/2 5. The point (-8, 15) is on the terminal side of an angle in standard position. Determine the exact values of the six trigonometric functions of the 7. Use the picture below to evaluate the following: a. sinθ = 1/5 b. cscθ = 5 c. tan(90° - θ) = 2√6 d. sec(90° - θ) = 5 8. Simplify each expression. a. (sin²θ + cos²θ)/(1 - cos²θ) = csc²θ b. (sinθ + cosθ)² + (sinθ - cosθ)² = 2 c. (tan²θ)/(secθ + 1) + 1 = secθ 9. In a sightseeing boat near the base of the Horseshoe Falls at Niagara Falls, a passenger estimates the angle of elevation to the top of the Falls to be   30°. If the Horseshoe Falls are 173 feet high, what is the distance from the boat to the base of the Falls?   tan(30°) = 173/x x = 173√3 ≈ 299.64 ft 10. In traveling across flat land, you notice a mountain directly in front of you. Its angle of elevation to the peak is 4°.   After you drive 17 miles closer to the mountain, the angle of elevation is 10°. Approximate the height of the mountain. tan(4°) = h/(x+17) tan(10°) = h/x (x+17)tan(4°) = h x*tan(10°) = h 17tan(4°) + x*tan(4°) = x*tan(10°) 1.19 = 0.106x x ≈ 11.2 h = x*tan(10°) = 1.96 Height ≈ 11.2 miles 11. At a point 150 feet from the base of a building, the angle of elevation to the bottom of a smokestack on the roof of the building is 28°,   and the angle of elevation to the top of the smokestack is 41°. Find the height of the smokestack. tan(28°) = y/150 tan(41°) = w/150 x = w - y w = 130.39 y = 79.76 x = 50.63 ft 12. A ship leaves port at 3:00 pm on a bearing of N 50° E. If the ship sails at 25 knots, find how many nautical miles east the ship has traveled from its point of departure at 7:00 p.m.   sin(50°) = e/100 e = 76.6 nm 13. A ship is 50 miles west and 22 miles south of port. Find the bearing the captain must take to sail back to port. tan(θ) = 50/22 θ = 66.3° N 66.3° E 14. A circle has a radius of 7 inches. Find the length of the arc intercepted by a central angle of 240°. S = rθ = 7 * (240°/180°) * π = 29.32 in 15. The circular blade on a saw has a radius of 4 inches and rotates at 2400 revolutions per minute. a. Find the angular speed in radians per second.   2400 rev/min * 2π rad/rev * 1 min/60 sec = 80π rad/sec ≈ 251.3 rad/min b. Find the linear speed of a blade tip in miles per hour. 2400 rev/min * 60 min/hr * 8in/rev * 1 ft/12 in * 1 mile/5280 ft = 57.1 mph 16. A satellite in a circular orbit 1125 km above a planet makes one complete revolution every 120 minutes. Assuming that the planet is a sphere of radius 6400 km, find the linear speed of the satellite in kilometers per minute. Round your answer to the nearest whole number.   r = 1125 + 6400 = 7525   km circumference / time = 2π(7525) / 120 ≈ 394 km/min 17. Find two solutions of the equation. Give your answers in degrees (0° ≤ θ < 360°) and in radians (0 ≤ θ < 2π) a. cotθ = -√3/3 θ = 2π/3 or 120° θ = 5π/3 or 300° b. secθ = √2 θ = π/4 or 45° θ = 7π/4 or 315°