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Question Number 134801 by deleteduser12 last updated on 07/Mar/21
Answered by EDWIN88 last updated on 07/Mar/21
13θ=π⇒cos13θ=−1let:z=cosθcos2θcos3θcos4θcos5θcos6θ2zsinθ=sin2θcos2θcos3θcos4θcos5θcos6θ4zsinθ=sin4θcos3θcos4θcos5θcos6θ8zsinθ=sin8θcos3θcos5θcos6θnotesin8θ=sin(8π13)=sin(π−5π13)=sin5θ8zsinθ=sin5θcos3θcos5θcos6θ16zsinθ=sin10θcos3θcos6θnotesin10θ=sin(10π13)=sin(π−3π13)=sin3θ16zsinθ=sin3θcos3θcos6θ32zsinθ=sin6θcos6θ64zsinθ=sin12θ⇒z=sin(12π13)64sin(π13)=164=126
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