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Honors Precalculus Chapter 5 Review Sheet 1) Find the exact, simplified values of cos θ and cot θ if csc θ = -10/3 and tan θ > 0. Name ____________________________ Period ______ 5 2) Find the exact, simplified values of sin θ and tan θ if cos θ = 5/11 and cot θ < 0. Use a half-angle formula to find the exact, simplified value of the expression: 3) tan 112.5o = 4) o sin 15 = Write the following expression as the sine, cosine, or tangent of an angle: 5) (cos 3π/8)(cos π/10) + (sin 3π/8)(sin π/10) = 6) o o tan 43 – tan 12 o o 1 + (tan 12 )(tan 43 ) Simplify the following expressions. Correct steps must be shown for full credit. 7) cot2x csc x - 1 2 8) tan x - sec2x cot2x = Honors Precalculus Chapter 5 Review Sheet -2- Find all solutions to the equation on the interval [0, 2π): 9) sin x + tan x – cos x = 1 Prove the following trigonometric identity: 11) 4 sin2x cos2x – 1 = - cos2(2x) 10) 2 cos2x + sin x – 1 = 0 Honors Precalculus Chapter 5 Review Sheet -3- Prove each trigonometric identity: 12) 13) 3 – tan2x cos2x – cos2x = 2 sin x tan x 2 cos x - 1 cos3x = -sec x Honors Precalculus Chapter 5 Review Sheet -4- 14) Find the area of a triangle with side lengths of 23 in, 34 in, and 37 in: 15) Solve the following triangle: C 28 mm B 64o A 16) 52.6 mm Solve the following triangle: F 7.3 cm D 9.2 cm 43o E Honors Precalculus Chapter 5 Review Sheet 17) -5- Solve the following triangle: R 75.6” Q 18) 33o 48.8” S A flagpole is attached to the top of a skyscraper. From a point on the ground 800 feet from the base of this building, the angle of elevation to the top of the flagpole is 44.8o and the angle of elevation to the bottom of the flagpole is 42.7o. Find the height of the flagpole: Honors Precalculus Chapter 5 Review Sheet -6- 19) An airplane travels due west from Boston at 750 MPH. After one hour, the plane turns and flies due northwest at the same speed for 20 minutes. How far from Boston is the plane at that point? 20) A triangle has side lengths of 29 cm 36 cm, and 45 cm. Find the measure of its largest angle: ANSWERS: 1) cos θ = - 91/10 7) 1 + csc x cot θ = 91/3 16) e = 6.300 cm mD = 52.205o mF = 84.795o 8) -tan4x 2) sin θ = -4 6/11 tan θ = -4 6/5 3) -1 - 2 or 4) ½ 2– 3 5) cos (11π/40) o 6) tan 31 9) x = π/4, π, 5π/4 2+ 2 2- 2 or 3+2 2 17) m S = 57.538o m R = 89.462o r = 89.597 inches or 10) x = π/2, 7π/6, 11π/6 14) 382.936 in2 15) mB = 28.584o mC = 87.416o c = 58.463 mm mS = 122.462o mR = 24.538o r = 37.210 inches 18) 56.216 feet 19) 943.486 miles 20) 86.925o