Download Math 110 Review: Identities, Trig Equations, & Angle Addition Formulas and more Exams Trigonometry in PDF only on Docsity! Math 110 3–1–2009 Review Problems for Test 3 These problems are meant to help you study. The presence of a problem on this sheet does not imply that there will be a similar problem on the test. And the absence of a problem from this sheet does not imply that the test does not have such a problem. 1. Prove the identity 1 tan x − 1 cotx = 2 cot 2x. 2. Does (sin x)2 − 5 sinx + 6 = 0 have any solutions in the interval 0 ≤ x ≤ 2π? 3. Prove the identity secx sin x csc x cosx = (tanx)2. 4. Find all solutions to cos 2θ = − 1√ 2 in the interval 0 ≤ θ ≤ 2π. 5. Find all solutions to 2 tanx sin x − tan x = 0 in the interval 0◦ ≤ x ≤ 360◦. 6. Use the angle addition formulas for sine and cosine to prove tan(a + b) = tan a + tan b 1 − tana tan b . Solutions to the Review Problems for Test 3 1. Prove the identity 1 tan x − 1 cotx = 2 cot 2x. 1 tan x − 1 cotx = 1 sinx cosx − 1cosx sinx = cosx sinx − sin x cosx = cosx cosx · cosx sinx − sin x sin x · sinx cosx = (cosx)2 sin x cosx − (sin x) 2 sin x cosx = (cos x)2 − (sin x)2 sin x cosx = 2 · (cosx) 2 − (sin x)2 2 sinx cosx = 2 cos 2x sin 2x = 2 cot2x. 2. Does (sin x)2 − 5 sinx + 6 = 0 have any solutions in the interval 0 ≤ x ≤ 2π? (sin x)2 − 5 sinx + 6 = 0, (sin x − 2)(sin x − 3) = 0, sin x = 2 or sinx = 3. Since sin never produces values greater than 1, sinx = 2 and sinx = 3 have no solutions. 1 3. Prove the identity secx sin x csc x cosx = (tanx)2. sec x sin x cscx cosx = 1 cosx · sinx 1 sin x · cosx = (sin x)2 (cosx)2 = ( sin x cosx )2 = (tanx)2. 4. Find all solutions to cos 2θ = − 1√ 2 in the interval 0 ≤ θ ≤ 2π. I know that cos 3π 4 = − 1√ 2 and cos 5π 4 = − 1√ 2 . 3 /4π 5 /4π -1 2 2 3π 4 + 2π = 11π 4 and 5π 4 + 2π = 13π 4 , so cos 11π 4 = − 1√ 2 and cos 13π 4 = − 1√ 2 as well. 2θ = 3π 4 gives θ = 3π 8 , 2θ = 5π 4 gives θ = 5π 8 , 2θ = 11π 4 gives θ = 11π 8 , 2θ = 13π 4 gives θ = 13π 8 . The solutions in the interval 0 ≤ θ ≤ 2π are 3π 8 , 5π 8 , 11π 8 , and 13π 8 . 5. Find all solutions to 2 tanx sin x − tan x = 0 in the interval 0◦ ≤ x ≤ 360◦. 2 tanx sin x − tan x = 0, (tanx)(2 sin x − 1) = 0. tan x = 0 has the solutions x = 0◦, x = 180◦, and x = 360◦. 2 sinx − 1 = 0 gives sin x = 1 2 . The solutions are x = 30◦ and x = 150◦. 150o 30o 1 2 2 1 2
