Question

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

• A. $6\le n\le 8$
• B. $4<n\le 8$
• C. $4\le n\le 8$
• D. $4<n<8$

Answer 1

The correct option is D

$4<n<8$

Step 1: Determine the range of $\theta$.

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

Squaring on both sides we get,

$$\begin{gathered} {\left(\sin \left(\frac{\pi }{2n}\right)+\cos \left(\frac{\pi }{2n}\right)\right)}^2={\left(\frac{\sqrt{n}}{2}\right)}^2 \\ \Rightarrow {\sin }^2\left(\frac{\pi }{2n}\right)+{\cos }^2\left(\frac{\pi }{2n}\right)+2\sin \left(\frac{\pi }{2n}\right)\cos \left(\frac{\pi }{2n}\right)=\frac{n}{4} \\ \Rightarrow 1+\sin \left(2\frac{\pi }{2n}\right)=\frac{n}{4}\because \left(2\sin \left(A\right)\cos \left(A\right)=\sin \left(2A\right)\right) \end{gathered}$$

$\Rightarrow$ $\sin \left(\frac{\pi }{n}\right)=\frac{n}{4}-1$

Now the value of $\sin \left(\theta \right)$ ranges in between $\left[-1,1\right]$ that is,

$-1\le \sin \left(\theta \right)\le 1$

Step 2: Determine the range of $n$

Now it is given that $n$ is positive, hence

$$\begin{gathered} 0\le \sin \left(\frac{\pi }{n}\right)\le 1 \\ \Rightarrow 0\le \left(\frac{n}{4}-1\right)\le 1 \\ \Rightarrow 1\le \frac{n}{4}\le 2 \\ \Rightarrow 4\le n\le 8 \end{gathered}$$

Now check the values at extreme points, we get

For $n=4$, $\sin \left(\frac{\pi }{4}\right)=\frac{4}{4}-1=0$ which is not true.

Also, for $n=8$, $\sin \left(\frac{\pi }{8}\right)=\frac{8}{4}-1=1$ which is also not true. Therefore $4<n<8$

Hence, option D is the correct answer.