Stitz-Zeager_College_Algebra_e-book

Solving for x gives x 5 k or x 7 k for integers k

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Unformatted text preview: ) + A cos(ωx) sin(φ) + B Once again, we may take ω = 2 and B = 0 so that cos(2x) − √ 3 sin(2x) = A sin(2x) cos(φ) + A cos(2x) sin(φ) √ We equate9 the coefficients of cos(2x) on either side and get A sin(φ) = 1 and A cos(φ) = − 3. Using A2 cos2 (φ) + A2 sin2 (φ) = A2 as before, we get A = ±2, and again we choose A = 2. √ √ This means 2 sin(φ) = 1, or sin(φ) = 1 , and 2 cos(φ) = − 3, which means cos(φ) = − 23 . 2 π π One such angle which meets these criteria is φ = 56 . Hence, we have f (x) = 2 sin 2x + 56 . Checking our work analytically, we have f (x) = 2 sin 2x + 5π 6 = 2 sin(2x) cos 5π 6 3 2 √ + cos(2x) sin + cos(2x) = 2 sin(2x) − √ = cos(2x) − 3 sin(2x) 5π 6 1 2 Graphing the three formulas for f (x) result in the identical curve, verifying our analytic work. It is important to note that in order for the technique presented in Example 10.5.3 to fit a function into one of the forms in Theorem √ 10.23, the arguments of the cosine and sine√ function much match. That is, while f (x) = cos(2x)...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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