Question

show that $\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta }}}=2\cos \theta$, $0<\theta <\frac{\pi }{8}$

Answer 1

According to the problem:

$\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta }}}$

implies that,

$=\sqrt{2+\sqrt{2+\sqrt{2+\left(1+\cos 8\theta \right)}}}$

implies that,

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

implies that,

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

implies that,

$=\sqrt{2+\left(2\sqrt{1+\cos 4\theta }\right)}$

implies that,

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

implies that,

$=\sqrt{2\left(1+\cos 2\theta \right)}$

implies that,

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

Hence the result will be,

$=2\cos \theta$