Question

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

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

Answer 1

Answer: (A)

n=6

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

Upon squaring both side we get

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

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

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

As $-1\le \sin \frac{\pi }{n}\le 1$

It is given $n$ is a positive integer.So,

$0\le \sin \frac{\pi }{n}\le 1$

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

$0\le n-4\le 4$

$4\le n\le 8$

This means when $n=4$ , $\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ , but according to solution $\sin \frac{\pi }{4}=0$ which can't be possible so $n=4$ does not satisfy.

Similarly when $n=8$ , the value of $\sin \frac{\pi }{8}\ne 1$ .Thus $n=8$ also not satisfied .

when $n=6$ ,$\sin \frac{\pi }{6}=\frac{1}{2}$ which satisfy in the condition ,

So answer will be $n=6$