Question

Let $n$ be a positive integer such that $\sin \frac{\pi }{2^n}+\cos \frac{\pi }{2^n}=\frac{\sqrt{n}}{2}$ , then

• A. $5\le n<8$
• B. $5<n<9$
• C. $4\le n<8$
• D. $4<n\le 8$

Answer 1

The correct option is D $4<n\le 8$

$\sin \left(\frac{\pi }{2^n}\right)+\cos \left(\frac{\pi }{2^n}\right)=\frac{\sqrt{n}}{2},n\ge 0$

Squaring on both sides,

${\sin }^2\left(\frac{\pi }{2^n}\right)+{\cos }^2\left(\frac{\pi }{2^n}\right)+\sin \left(\frac{2\pi }{2^n}\right)=\frac{n}{4}$

$1+\sin \left(\frac{\pi }{2^{n-1}}\right)=\frac{n}{4}$

$\sin \left(\frac{\pi }{2^{n-1}}\right)=\frac{n-4}{4}$
$0<\frac{n-4}{4}\le 1$

$0<n-4\le 4$

$\Rightarrow 4<n\le 8$