Question

The number of all the possible integral values of $n>2$ such that $\sin \frac{\pi }{2n}+\cos \frac{\pi }{2n}=\frac{\sqrt{n}}{2}$ is

• A. 5
• B. 4
• C. 3
• D. infinity

Answer 1

Answer: (C)

Given,
$\sin \frac{\pi }{2n}+\cos \frac{\pi }{2n}=\frac{\sqrt{n}}{2}\dots (i)$
Squaring both sides
${\left(\sin \frac{\pi }{2n}+\cos \frac{\pi }{2n}\right)}^2=\frac{n}{4}$
${\sin }^2\frac{\pi }{2n}+{\cos }^2\frac{\pi }{2n}+2\sin \frac{\pi }{2n}\cos \frac{\pi }{2n}=\frac{n}{4}$
$\Rightarrow 1+\sin \frac{\pi }{n}=\frac{n}{4}$
$\Rightarrow \sin \frac{\pi }{n}=\frac{n}{4}-1$
$\Rightarrow \sin \frac{\pi }{n}=\frac{n-4}{4}$
As, $\sin \frac{\pi }{n}>0,n>2$
So, $\frac{n-4}{4}>0$
$\Rightarrow n>4\dots (ii)$
Also, $\sin \left(\frac{\pi }{2n}\right)+\cos \frac{\pi }{2n}=\sqrt{2}\sin \left(\frac{\pi }{4}+\frac{\pi }{2n}\right)$.
From Eq. (i)
$\sqrt{2}\sin \left(\frac{\pi }{4}+\frac{\pi }{2n}\right)=\frac{\sqrt{n}}{2}\sin \left(\frac{\pi }{4}+\frac{\pi }{2n}\right)=\frac{\sqrt{n}}{2\sqrt{2}}$
Since, $\sin \left(\frac{\pi }{4}+\frac{\pi }{2n}\right)<1,\forall n>2$
We get, $\frac{\sqrt{n}}{2\sqrt{2}}<1$
$\Rightarrow n<8\dots (iii)$
From Eqs. (ii) and (iii) $4<n<8$
So, $n=5,6,7$
Hence, number of integral value of $n$ is 3