Question

If $0<\theta <\frac{\pi }{8}$, show that $\sqrt{2+\sqrt{2+\sqrt{2+2cos4\theta }}}=2cos\left(\theta /2\right).$

Answer 1

b'

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

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

$=\sqrt{2+\sqrt{2+\sqrt{2(1+2{\cos }^{2}2\theta -1)}}}$

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

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

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

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

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

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

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

$=2\cos \frac{\theta }{2}$

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