Question

if $\sqrt{2}cos\theta$ = $\sqrt{\frac{2+\sqrt{2+\sqrt{2}}}{2}}$ then the value of $\theta$ is

• A. $\frac{\pi }{16}$
• B. $\frac{\pi }{32}$
• C. $\frac{\pi }{64}$
• D. $\frac{\pi }{128}$

Answer 1

The correct option is A $\frac{\pi }{16}$

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

Squaring on both sides

$2{\cos }^2\theta =1+\frac{\sqrt{2+\sqrt{2}}}{2}$

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

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

Squaring on both sides

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

$⟹2{\cos }^{2}2\theta =1+\frac{1}{\sqrt{2}}$

$⟹\cos 4\theta =\frac{1}{\sqrt{2}}$

$⟹4\theta =\frac{\pi }{4}$

$⟹\theta =\frac{\pi }{16}$