Question

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

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

Answer 1

Answer: (C)

$n=6$

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

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

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

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

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

So, that $4<n<8$.

By actual verification, we find that only $n=6$. satisfies the given relation.