Question

If $n$ is a positive integer such $\sin \frac{\pi }{2^n}+\cos \frac{\pi }{2^n}=\frac{\sqrt{n}}{2}$ then

Answer 1

$\sin \frac{\pi }{2^n}+\cos \frac{\pi }{2^n}=\frac{\sqrt{n}}{2}$

Squaring both sides,we get

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

$\Rightarrow {\sin }^2\frac{\pi }{2^n}+{\cos }^2\frac{\pi }{2^n}+2\sin \frac{\pi }{2^n}\cos \frac{\pi }{2^n}=\frac{n}{4}$

$\Rightarrow 1+\sin \frac{2\pi }{2^n}=\frac{n}{4}$

$\Rightarrow \sin \frac{\pi }{2^{n-1}}=\frac{n}{4}-1=\frac{n-4}{4}$

Since $\sin \theta \le 1\Rightarrow \frac{n-4}{4}\ge 0$ ......$\left(1\right)$

where $n$ is a positive integer$\Rightarrow n-1\ge 0$

$\Rightarrow 2^{n-1}\ge 1$

$\sin \left(\frac{\pi }{2^{n-1}}\right)\ge 0$

$\Rightarrow \frac{n-4}{4}\ge 0$ .......$\left(2\right)$

$\therefore 0\le \frac{n-4}{4}\le 1$

$\Rightarrow 0\le n-4\le 4$

$\therefore 4\le n\le 8$