Q. Let $n$ be a positive integer such that $\sin{\frac{\pi}{2n} + \cos{\frac{\pi}{2n} = \frac{\sqrt{n}}{2}}}$. Then
A) $n = 6$
B) $n = 1,\ 2,\ 3\ ,\ldots,8$
C) $n = 5$
D) None of these
Explanation
(a)
$\because\ \sin{\left( \ \frac{\pi}{2n} \right) + {cos(\ }{\frac{\pi}{2n})\ = \frac{\sqrt{n}}{2}}}$
On squaring both sides, we get
$\sin^{2}{\left( \ \frac{\pi}{2n} \right) + \cos^{2}{\left( \frac{\pi}{2n} \right) + \sin{\left( \ \frac{\pi}{n} \right) = \frac{n}{4}}\ }}$
$\Rightarrow \ \sin{\left( \ \frac{\pi}{n} \right) = \frac{n}{4} - 1}$
$\Rightarrow \ \sin{\left( \ \frac{\pi}{n} \right) = \frac{n - 4}{4}}$
$\Rightarrow \ n = 6$ only
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.