Stitz-Zeager_College_Algebra_e-book

Using the above guidelines we can comfortably solve

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Unformatted text preview: . What we have here is a system of nonlinear equations! We can temporarily eliminate the dependence on φ by using the 2 Pythagorean Identity. We know cos2 (φ) + sin2 (φ) = 1, so multiplying this by A√ gives √ 2 cos2 (φ)+A2 sin2 (φ) = A2 . Since A cos(φ) = 1 and A sin(φ) = 2 = 12 +( 3)2 = A 3, we get√ A 4 or A = ±2. Choosing A = 2, we have 2 cos(φ) = 1 and 2 sin(φ) = 3 or, after some √ 1 rearrangement, cos(φ) = 2 and sin(φ) = 23 . One such angle φ which satisfies this criteria is φ = π . Hence, one way to write f (x) as a sinusoid is f (x) = 2 cos 2x + π . We can easily 3 3 check our answer using the sum formula for cosine f (x) = 2 cos 2x + π 3 = 2 cos(2x) cos = 2 cos(2x) = cos(2x) − 8 1 2 √ π 3 − sin(2x) sin − sin(2x) √ π 3 3 2 3 sin(2x) This should remind you of equation coefficients of like powers of x in Section 8.6. 10.5 Graphs of the Trigonometric Functions 681 2. Proceeding as before, we equate f (x) = cos(2x) − S (x) = A sin(ωx + φ) + B to get cos(2x) − √ √ 3 sin(2x) with the expanded form of 3 sin(2x) = A sin(ωx) cos(...
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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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