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Question

Let z=cosθ+i sinθ. The value of 15m=1Im(z2m1) at θ=2

A
1sin2
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B
13sin2
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C
12sin2
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D
14sin2
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Solution

The correct option is D 14sin2
z=cosθ+i sinθ=eiθ15m=1Im(z2m1)=15m=1Im(eiθ)2m1=15m=1Im ei(2m1)θ=sin θ+sin 3θ+sin 5θ+...+sin 29θ=sin(θ+ 29θ2)sin(15×2θ2)sin(2θ2)=sin 15θ.sin 15θsin θ=14 sin2

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