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Question Number 43543 by peter frank last updated on 11/Sep/18
prove that  (√(2+(√(2+(√(2+2cos 8θ))))))  2cos θ
$${prove}\:{that}\:\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\mathrm{2cos}\:\mathrm{8}\theta}}} \\ $$$$\mathrm{2cos}\:\theta \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Sep/18
(√(2+(√(2+(√(2+2cos8θ))))))  =(√(2+(√(2+(√(2(1+cos8θ))) ))))  =(√(2+(√(2+(√(4cos^2 4θ))))))  =(√(2+(√(2+2cos4θ))))  =(√(2+(√(2(1+cos4θ)))))  .=(√(2+(√(4cos^2 2θ))))  =(√(2+2cos2θ))  =(√(2(1+cos2θ)))  =(√(4cos^2 θ))  =2cosθ
$$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\mathrm{2}{cos}\mathrm{8}\theta}}} \\ $$$$=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\mathrm{8}\theta\right)}\:}} \\ $$$$=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{4}{cos}^{\mathrm{2}} \mathrm{4}\theta}}} \\ $$$$=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\mathrm{2}{cos}\mathrm{4}\theta}} \\ $$$$=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\mathrm{4}\theta\right)}} \\ $$$$.=\sqrt{\mathrm{2}+\sqrt{\mathrm{4}{cos}^{\mathrm{2}} \mathrm{2}\theta}} \\ $$$$=\sqrt{\mathrm{2}+\mathrm{2}{cos}\mathrm{2}\theta} \\ $$$$=\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\mathrm{2}\theta\right)} \\ $$$$=\sqrt{\mathrm{4}{cos}^{\mathrm{2}} \theta} \\ $$$$=\mathrm{2}{cos}\theta \\ $$$$ \\ $$

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