Q. $\left( 1 + \cos\frac{\pi}{8} \right)\left( 1 + \cos\frac{3\pi}{8} \right)\left( 1 + \cos\frac{5\pi}{8} \right)\left( 1 + \cos\frac{7\pi}{8} \right)$ is equal to

A) $\frac{1}{2}$
B) $\cos\frac{\pi}{8}$
C) $\frac{1}{8}$
D) $\frac{1 + \sqrt{2}}{2\sqrt{2}}$
Explanation

(c)
$\left( 1 + \cos\frac{\pi}{8} \right)\left( 1 + \cos\frac{3\pi}{8} \right)\left( 1 + \cos\frac{5\pi}{8} \right)\left( 1 + \cos\frac{7\pi}{8} \right)$
$= \left( 1 + \cos\frac{\pi}{8} \right)\left( 1 + \cos\frac{3\pi}{8} \right)\left( 1 - \cos\frac{3\pi}{8} \right) \times \left( 1 - \cos\frac{\pi}{8} \right)$
$= \left( 1 - \cos^{2}\frac{\pi}{8} \right)\left( 1 - \cos^{2}\frac{3\pi}{8} \right)$
$= {\sin^{2}{\frac{\pi}{8}.\sin^{2}{\frac{3\pi}{8} = \frac{1}{4}\left\lbrack 2\sin{\frac{\pi}{8}\sin\frac{3\pi}{8}} \right\rbrack}}}^{2}$
$= \frac{1}{4}\left\lbrack \cos{\frac{\pi}{4} - \cos\frac{\pi}{2}} \right\rbrack^{2} = \frac{1}{4}\left\lbrack \frac{1}{\sqrt{2}} - 0 \right\rbrack^{2} = \frac{1}{8}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.