Question

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

• A. $n=6$
• B. $n=1,2,3,....8$
• C. $n=5$
• D. $n=4$

Answer 1

The correct option is A $n=6$

$\sin \frac{\pi }{2n}+\cos \frac{\pi }{2n}=\frac{\sqrt{n}}{2}\Rightarrow \frac{1}{\sqrt{2}}\sin \frac{\pi }{2n}+\frac{1}{\sqrt{2}}\cos \frac{\pi }{2n}=\frac{\sqrt{n}}{2\sqrt{2}}\Rightarrow \cos \frac{\pi }{4}\sin \frac{\pi }{2n}+\sin \frac{\pi }{4}\cos \frac{\pi }{2n}=\frac{\sqrt{n}}{2\sqrt{2}}\Rightarrow \sin (\frac{\pi }{2n}+\frac{\pi }{4})=\frac{\sqrt{n}}{2\sqrt{2}}\Rightarrow (\frac{\pi }{2n}+\frac{\pi }{4})={\sin }^{-1}\frac{\sqrt{n}}{2\sqrt{2}}$
As $-1\le \frac{\sqrt{n}}{2\sqrt{2}}\le 1\Rightarrow -2\sqrt{2}\le \sqrt{n}\le 2\sqrt{2}\Rightarrow -8\le n\le 8$
Only $n=6$ satisfy