Question

Let $n$ be a fixed positive integer such that $\sin \left(\frac{\pi }{2n}\right)+\cos \left(\frac{\pi }{2n}\right)=\frac{\sqrt{n}}{2}$, then

• A. $n=4$
• B. $n=5$
• C. $n=6$
• D. none of these

Answer 1

Answer: (B), (C)

The correct options are
B $n=5$
C $n=6$

We have, $\sin \left(\frac{\pi }{2n}\right)+\cos \left(\frac{\pi }{2n}\right)=\sqrt{2}\sin (\frac{\pi }{4}+\frac{\pi }{2n})$

$\Rightarrow \frac{\sqrt{n}}{2}=\sqrt{2}\sin (\frac{\pi }{4}+\frac{\pi }{2n})$

So for $n>1,\frac{\sqrt{n}}{2\sqrt{2}}=\sin (\frac{\pi }{4}+\frac{\pi }{2n})>\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$

Thus $n>4$

Since $\sin (\frac{\pi }{4}+\frac{\pi }{2n})<1$ for all $n>2$,

we get $\frac{\sqrt{n}}{2\sqrt{2}}<1\Rightarrow n<8$

$\therefore 4<n<8$