2 a z 12cis 6 6i 3 c z 2cis 78 2 2

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Unformatted text preview: or r = −r . If r = r , then θ = θ + 2πk , so one possibility is that an intersection point P has a representation (r, θ) which satisfies r = 3 and another representation (r, θ + 2πk ) for some integer, k which satisfies r = 6 cos(2θ). At this point, we replace every occurrence θ in the equation r = 6 cos(2θ) with (θ +2πk ) and then see if, by equating the resulting expressions for r, we get any more solutions for θ.13 Since cos(2(θ + 2πk )) = cos(2θ + 4πk ) = cos(2θ) for every integer k , however, the equation r = 6 cos(2(θ + 2πk )) reduces to the same equation we had before, r = 6 cos(2θ), which means we get no additional solutions. Moving on to the case where r = −r , we have that θ = θ + (2k + 1)π for integers k . We look to see if we can find points P which have a representation (r, θ) that satisfies r = 3 and another, 12 See Example 11.5.2 number 3. The authors have chosen to replace θ with θ + 2πk in the equation r = 6 cos(2θ) for illustration purposes only. We could have just...
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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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