Stitz-Zeager_College_Algebra_e-book

62 2 3 3 3 3 6 2 w0 2cis 12 2 i 2 i 2 3 4 4 2 2 w1

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Unformatted text preview: as easily chosen to do this substitution in the equation r = 3. Since there is no θ in r = 3, however, this case would reduce to the previous case instantly. The reader is encouraged to follow this latter procedure in the interests of efficiency. 13 812 Applications of Trigonometry (−r, θ + (2k + 1)π ), that satisfies r = 6 cos(2θ). To do this, we substitute14 (−r) for r and (θ + (2k + 1)π ) for θ in the equation r = 6 cos(2θ) and get −r = 6 cos(2(θ + (2k + 1)π )). Since cos(2(θ + (2k + 1)π )) = cos(2θ + (2k + 1)(2π )) = cos(2θ) for all integers k , the equation −r = 6 cos(2(θ + (2k + 1)π )) reduces to −r = 6 cos(2θ), or r = −6 cos(2θ). Coupling this 1 π equation with r = 3 gives −6 cos(2θ) = 3 or cos(2θ) = − 2 . We get θ = π + πk or θ = 23 + πk . 3 From these solutions, we obtain15 the remaining four intersection points with representations π π π −3, π , −3, 23 , −3, 43 and −3, 53 , which we can readily check graphically. 3 θ θ 4. As usual, we begin by graphing r = 3 sin 2 and...
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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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