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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 ⇒cos 13θ = −1  let : z = cos θ cos 2θ cos 3θ cos 4θ cos 5θ cos 6θ  2z sin θ = sin 2θ cos 2θ cos 3θ cos 4θ cos 5θ cos 6θ  4z sin θ = sin 4θ cos 3θ cos 4θ cos 5θ cos 6θ  8z sin θ = sin 8θ cos 3θ cos 5θ cos 6θ  note sin 8θ = sin (((8π)/(13)))=sin (π−((5π)/(13)))=sin 5θ  8z sin θ = sin 5θ cos 3θ cos 5θ cos 6θ  16z sin θ = sin 10θ cos 3θ cos 6θ  note sin 10θ=sin (((10π)/(13)))=sin (π−((3π)/(13)))=sin 3θ  16z sin θ = sin 3θ cos 3θ cos 6θ  32z sin θ = sin 6θ cos 6θ   64z sin θ = sin 12θ ⇒z = ((sin (((12π)/(13))))/(64 sin ((π/(13))))) = (1/(64))=(1/2^6 )

13θ=πcos13θ=1let:z=cosθcos2θcos3θcos4θcos5θcos6θ2zsinθ=sin2θcos2θcos3θcos4θcos5θcos6θ4zsinθ=sin4θcos3θcos4θcos5θcos6θ8zsinθ=sin8θcos3θcos5θcos6θnotesin8θ=sin(8π13)=sin(π5π13)=sin5θ8zsinθ=sin5θcos3θcos5θcos6θ16zsinθ=sin10θcos3θcos6θnotesin10θ=sin(10π13)=sin(π3π13)=sin3θ16zsinθ=sin3θcos3θcos6θ32zsinθ=sin6θcos6θ64zsinθ=sin12θz=sin(12π13)64sin(π13)=164=126

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