Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: −r = 3 cos then sin (2k+1)π 2 1 2 (2k+1)π 2 cos (2k+1)π 2 θ 2 + (2k+1)π 2 π 2 θ 2 sin (2k+1)π 2 (2k+1)π = ±1 for integers k . 2 θ as r = ±3 sin 2 . If we choose k = 0, = 3 cos 1 [θ + (2k + 1)π ] in this case 2 = 0 and sin [θ + (2k + 1)π ] can be rewritten = sin − sin = 1, and the equation −r θ 2 θ reduces to −r = −3 sin , or r = 3 sin 2 which is the other equation under consideration! θ What this means is that if a polar representation (r, θ) for the point P satisfies r = 3 sin( 2 ), θ then the representation (−r, θ + π ) for P automatically satisfies r = 3 cos 2 . Hence the θ θ equations r = 3 sin( 2 ) and r = 3 cos( 2 ) determine the same set of points in the plane. Our work in Example 11.5.3 justifies the following. Guidelines for Finding Points of Intersection of Graphs of Polar Equations To find the points of intersection of the graphs of two polar equations E1 and E2 : • Sketch the graphs of E1 and E2 . Check to see if the curves intersect at t...
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