Question

Prove that $\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta }}}=2\cos \theta$

Answer 1

L.H.S$\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta }}}$

$=\sqrt{2+\sqrt{2+\sqrt{2(1+\cos 8\theta )}}}[\because \cos 8\theta =2{\cos }^{2}4\theta -1\Rightarrow \cos 8\theta +1=2{\cos }^{2}4\theta ]$

$=\sqrt{2+\sqrt{2+\sqrt{4{\cos }^{2}4\theta }}}$

$=\sqrt{2+\sqrt{2+2\cos 4\theta }}$

$=\sqrt{2+\sqrt{2(1+\cos 4\theta )}}[\because \cos 4\theta =2{\cos }^{2}2\theta -1\Rightarrow \cos 4\theta +1=2{\cos }^{2}2\theta ]$

$=\sqrt{2+\sqrt{4{\cos }^2\left(2\theta \right)}}$

$=\sqrt{2+2\cos \left(2\theta \right)}$

$=\sqrt{2(1+\cos 2\theta )}[\because \cos 2\theta =2{\cos }^2\theta -1\Rightarrow \cos 2\theta +1=2{\cos }^2\theta ]$

$=\sqrt{2\left(2{\cos }^2\theta \right)}$

$=2\cos \theta =$R.H.S