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Math 1613 - Trigonometry Homework #5 - 2007.09.18 Due Date - 2007.09.27 Solutions 1. Verify the following identities. a) cot(−t) cos(t) + sin(−t) = − csc(t) cos(t) cos(t) − sin(t) sin(t) 1 =− cos2 (t) + sin2 (t) sin(t) 1 =− sin(t) = − csc(t). cot(−t) cos(t) + sin(−t) = − b) cos(−x) 1 + sin(−x) = 1 − sin(−x) cos(x) cos(−x) cos(x) = 1 − sin(−x) 1 + sin(x) cos(x) 1 − sin(x) = · 1 + sin(x) 1 − sin(x) cos(x)(1 − sin(x)) = 1 − sin2 (x) cos(x)(1 − sin(x)) = cos2 (x) 1 − sin(x) = cos(x) 1 + sin(−x) = . cos(x) 1 c) log(tan(−t)) = log(cos(t)) − log(sin(t)) The left side does not equal the right side! 2. Complete the following statements. a) as t → π + , sin(t) → as t → π + , sin(t) → 0. b) as t → π − , cos(t) → as t → π − , cos(t) → −1. c) as t → π + , csc(t) → as t → π + , csc(t) → −∞. d) as t → − π2 − , tan(t) → π − as t → − , tan(t) → ∞. 2 2 3. Find the exact values of the six trigonometric functions of θ if θ is in standard position and P is on the terminal side. a) P (−3, 7) First, we note that r = √ 58. 3 7 7 cos(θ) = − √ , sin(θ) = √ , tan(θ) = − 3 √58 √ 58 3 58 58 , csc(θ) = , cot(θ) = − sec(θ) = − 3 7 7 b) P (−7, −3) First, we note that r = √ 58. 7 3 3 cos(θ) = − √ , sin(θ) = − √ , tan(θ) = 7 √ 58 √58 58 58 7 sec(θ) = − , csc(θ) = − , cot(θ) = 7 3 3 4. (Problem 36 page 136) The highest advertising sign in the world is a large letter I situated at the top of the 73-story First Interstate World Center in Los Angeles. At a distance of 200 feet from a point directly below the sign, the angle between the ground and the top of the sign is 78.87◦ . Approximate the height of the top of the sign. To solve this, we use the fact that the distance of 200 feet forms the base of a triangle, and the distance to the top of the sign forms the height of the right triangle (call the height y). The angle opposite is 78.87◦ , which gives y tan (78.87◦ ) = . 200 This gives y ≈ 1016.590716 feet. 3 5. (Problem 38 page 136) The phases of the moon can be described using the phase angle θ, determined by the sun, the moon, and the earth, as shown in the figure at the bottom of page 136. Because the moon orbits the earth, θ changes during the course of a month. The area of the region A of the moon, which appears illuminated to an observer on the earth, is given by A = 21 πR2 (1 + cos(θ)), where R = 1080 miles is the radius of the moon. Approximate A for the following positions of the moon. a) θ = 0◦ (full moon) 1 A = πR2 (1 + cos(0)) 2 1 = πR2 (1 + 1) 2 = πR2 = 1166400π mi2 ≈ 3664353.672 mi2 . b) θ = 180◦ (new moon) 1 A = πR2 (1 + cos(π)) 2 1 2 = πR (1 − 1) 2 = 0 mi2 . 4 c) θ = 90◦ (first quarter) π 1 2 A = πR (1 + cos ) 2 2 1 = πR2 (1 + 0) 2 1 2 = πR 2 = 583200π mi2 ≈ 1832176.836 mi2 . 5