Question

For any $n\in N$, the value of the expression $\sqrt{2+\sqrt{2+\sqrt{2+...\text{n times}}}}$

is

• A. $2\cos \left(\frac{\pi }{2^{n+1}}\right)$
• B. $\sin \left(\frac{\pi }{2^{n+1}}\right)$
• C. $2\cos \left(2^{n+1}\pi \right)$
• D. $2\cos \left(2^n\pi \right)$

Answer 1

Answer: (A)

$2\cos \left(\frac{\pi }{2^{n+1}}\right)$

Let $2\cos x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\mathrm{2...}}}}}$
$\Rightarrow 4{\cos }^{2}x=2+\sqrt{2+\sqrt{2+\sqrt{\mathrm{2...}}}}\Rightarrow 2\cos 2x=\sqrt{2+\sqrt{2+\sqrt{\mathrm{2...}}}}\Rightarrow 4{\cos }^{2}2x=2+\sqrt{2+\sqrt{2+\sqrt{\mathrm{2...}}}}\Rightarrow 2\cos 4x=\sqrt{2+\sqrt{2+\sqrt{\mathrm{2...}}}}...\Rightarrow 2\cos 2^{n}x=0now\Rightarrow \cos 2^{n}x=0\Rightarrow 2^{n}x=\frac{\pi }{2}\Rightarrow x=\frac{\pi }{2^{n+1}}$
Therefore
$2\cos \frac{\pi }{2^{n+1}}=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\mathrm{2...}}}}}$