Stitz-Zeager_College_Algebra_e-book

The component form of a vector is what ties these

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Unformatted text preview: 1 [θ + 2πk ] = 3 cos 2 + πk . Using the sum formula for 2 θ θ θ θ cosine, we expand 3 cos 2 + πk = 3 cos 2 cos(πk ) − 3 sin 2 sin (πk ) = ±3 cos 2 , since sin(πk ) = 0 for all integers k , and cos (πk ) = ±1 for all integers k . If k is an even integer, θ θ we get the same equation r = 3 cos 2 as before. If k is odd, we get r = −3 cos 2 . This θ θ θ latter expression for r leads to the equation 3 sin 2 = −3 cos 2 , or tan 2 = −1. Solving, we get θ = − π + 2πk for integers k , which gives the intersection point 2 √ 32 π 2 ,−2 . Next, θ we assume P has a representation (r, θ) which satisfies r = 3 sin 2 and a representation θ (−r, θ + (2k + 1)π ) which satisfies r = 3 cos 2 for some integer k . Substituting (−r) for θ 1 r and (θ + (2k + 1)π ) in for θ into r = 3 cos 2 gives −r = 3 cos 2 [θ + (2k + 1)π ] . Once again, we use the sum formula for cosine to get cos 1 2 [θ + (2k + 1)π ] = cos = cos θ 2 θ 2 = ± sin where the last equality is true since cos Hence,...
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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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