Stitz-Zeager_College_Algebra_e-book

For f x secx we restrict the domain to 0 2 2 y y 1 x

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Unformatted text preview: as usual, this is something we need for Calculus. Since sin(2θ) = 2 sin(θ) cos(θ), we need to write cos(θ) in terms of x to finish the problem. We substitute x = sin(θ) into the Pythagorean Identity, cos2 (θ) + sin2 (θ) = 1, √ to get cos2 (θ) + x2 = 1, or cos(θ) = ± 1 − x2 . Since − π ≤ θ ≤ π , cos(θ) ≥ 0, and thus 2 2 √ √ cos(θ) = 1 − x2 . Our final answer is sin(2θ) = 2 sin(θ) cos(θ) = 2x 1 − x2 . 3. We start with the right hand side of the identity and note that 1 + tan2 (θ) = sec2 (θ). From this point, we use the Reciprocal and Quotient Identities to rewrite tan(θ) and sec(θ) in terms of cos(θ) and sin(θ): 2 tan(θ) 1 + tan2 (θ) = 2 tan(θ) sec2 (θ) 2 = sin(θ) cos(θ) 1 cos2 (θ) =2 sin(θ) cos(θ) =2 sin(θ) cos($ $$θ ) cos(θ ) cos($ $$θ ) cos2 (θ) = 2 sin(θ) cos(θ) = sin(2θ) 4. In Theorem 10.17, one of the formulas for cos(2θ), namely cos(2θ) = 2 cos2 (θ) − 1, expresses cos(2θ) as a polynomial in terms of cos(θ). We are now asked to find such an identity for cos(3θ). Using the sum formula for cosine, we begin with cos(3θ) = cos(2θ + θ) = cos(2θ) cos(θ) − sin(2θ) sin(θ) Our ultimate goal is to express the right hand side in terms of cos(θ) only. We substitute cos(2θ) = 2 cos2 (θ) − 1 and sin(2θ) = 2 sin(θ) cos(θ) which yields cos(3θ) = cos(2θ) cos(θ) − sin(2θ) sin(θ) = 2 cos2 (θ) − 1 cos(θ) − (2 sin(θ) cos(θ)) sin(θ) = 2 cos3 (θ) − cos(θ) − 2 sin2 (θ) cos(θ) 10.4 T...
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