Let z=cosθ+isinθ. The value of ∑15m=1Im(z2m−1) 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 D14sin2∘ z=cosθ+isinθ=eiθ∴∑15m=1Im(z2m−1)=∑15m=1Im(eiθ)2m−1=∑15m=1Imei(2m−1)θ=sinθ+sin3θ+sin5θ+...+sin29θ=sin(θ‘+29θ2)sin(15×2θ2)sin(2θ2)=sin15θ.sin15θsinθ=14sin2∘