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Question Number 134801 by faysal 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 )
$$\mathrm{13}\theta\:=\:\pi\:\Rightarrow\mathrm{cos}\:\mathrm{13}\theta\:=\:−\mathrm{1} \\ $$$$\mathrm{let}\::\:\mathrm{z}\:=\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{4}\theta\:\mathrm{cos}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{2z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{4}\theta\:\mathrm{cos}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{4z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{4}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{4}\theta\:\mathrm{cos}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{8z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{8}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{note}\:\mathrm{sin}\:\mathrm{8}\theta\:=\:\mathrm{sin}\:\left(\frac{\mathrm{8}\pi}{\mathrm{13}}\right)=\mathrm{sin}\:\left(\pi−\frac{\mathrm{5}\pi}{\mathrm{13}}\right)=\mathrm{sin}\:\mathrm{5}\theta \\ $$$$\mathrm{8z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{16z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{10}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{note}\:\mathrm{sin}\:\mathrm{10}\theta=\mathrm{sin}\:\left(\frac{\mathrm{10}\pi}{\mathrm{13}}\right)=\mathrm{sin}\:\left(\pi−\frac{\mathrm{3}\pi}{\mathrm{13}}\right)=\mathrm{sin}\:\mathrm{3}\theta \\ $$$$\mathrm{16z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{32z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{6}\theta\:\mathrm{cos}\:\mathrm{6}\theta\: \\ $$$$\mathrm{64z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{12}\theta\:\Rightarrow\mathrm{z}\:=\:\frac{\mathrm{sin}\:\left(\frac{\mathrm{12}\pi}{\mathrm{13}}\right)}{\mathrm{64}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{13}}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{64}}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} } \\ $$

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