Stitz-Zeager_College_Algebra_e-book

From theorem 103 we know that since the point b x y

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Unformatted text preview: of y = cos(3x) and y = cos(5x) below bears this out. √ 6. In examining the equation sin(2x) = 3 cos(x), not only do we have different circular functions involved, namely sine and cosine, we also have different arguments to contend with, namely 2x and x. Using the identity sin(2x) = 2 sin(x) cos(x) makes all of the arguments the same and we proceed as we would solving any nonlinear equation – gather all of the nonzero terms on one side of the equation and factor. sin(2x) 2 sin(x) cos(x) √ 2 sin(x) cos(x) − 3 cos(x) √ cos(x)(2 sin(x) − 3) √ = √3 cos(x) = 3 cos(x) (Since sin(2x) = 2 sin(x) cos(x).) =0 =0 √ from which we get cos(x) = 0 or sin(x) = √ integers k . From sin(x) = 6 7 3 2, we get x = As always, experience is the greatest teacher here! As always, when in doubt, write it out! π 3 3 2. From cos(x) = 0, we obtain x = +2πk or x = 2π 3 π 2 + πk for +2πk for integers k . The answers 736 Foundations of Trigonometry π π which lie in [0, 2π ) are x = π , 32 , π and 23 . We graph y = sin(2x) and y = 2 3 after some careful zooming, verify...
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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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