Question

Find the value of $\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}$.

• A. $\cos \frac{\pi }{4}$
• B. $\cos \frac{\pi }{8}$
• C. $\cos \frac{\pi }{16}$
• D. $\cos \frac{\pi }{32}$

Answer 1

The correct option is D $\cos \frac{\pi }{32}$

Let $\cos x=\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}$
$\Rightarrow 4{\cos }^{2}x=2+\sqrt{2+\sqrt{2+\sqrt{2}}}\Rightarrow 2\cos 2x=\sqrt{2+\sqrt{2+\sqrt{2}}}\Rightarrow 4{\cos }^{2}2x=2+\sqrt{2+\sqrt{2}}\Rightarrow 2\cos 4x=\sqrt{2+\sqrt{2}}\Rightarrow 4{\cos }^{2}4x=2+\sqrt{2}\Rightarrow 2\cos 8x=\sqrt{2}\Rightarrow 4{\cos }^{2}8x=2\Rightarrow 2\cos 16x=0now\Rightarrow 2\cos 2^{4}x=0\Rightarrow 2^{4}x=\frac{\pi }{2}\therefore x=\frac{\pi }{2^5}$
Hence
$\cos \frac{\pi }{32}=\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}$