Question

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

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

Answer 1

The correct option is D $4\le n\le 8$

$\begin{array}{c}\sin \left(\frac{\pi }{2n}\right)+\cos \left(\frac{\pi }{2n}\right)=\frac{\sqrt{n}}{2}\\ \text{Squaring both sides}\\ {\sin }^2\frac{\pi }{2n}+{\cos }^2\frac{\pi }{2n}\ne 2\cos \frac{\pi }{2n}\sin \frac{\pi }{2n}=\frac{n}{4}\\ \sin \frac{\pi }{n}=\frac{n}{4}-1\\ \sin \frac{\pi }{n}=\frac{n-4}{4}\\ \text{Now, we know that}\\ \begin{array}{cc}-1⩽\sin \theta & ⩽1\\ \frac{n-4}{4}& ⩽1\\ n-4& ⩽4\\ n& ⩽8\end{array}\end{array}$