Question

The value of
$\left(1+cos \frac{\pi}{10}\right)\left(1+cos \frac{3\pi }{10}\right)\left(1+cos \frac{7\pi }{10}\right)\left(1+cos \frac{9\pi }{10}\right)$ is

• A. $\frac{1}{8}$
• B. $\frac{1}{16}$
• C. $\frac{1}{32}$
• D. None of these

Answer 1

Answer: (B)

The expression
$= \left(1+cos \frac{\pi}{10}\right)\left(1+cos \frac{3\pi }{10}\right)\left(1-cos \frac{3\pi }{10}\right)\left(1-cos \frac{\pi }{10}\right)$
$\left[\because cos \frac{7\pi }{10} = cos\left(\pi-\frac{3\pi }{10}\right) = -cos \frac{3\pi }{10}\,and\,cos \frac{9\pi }{10} = cos\left(\pi -\frac{\pi }{10}\right) =- cos \frac{\pi }{10}\right]$
$= \left(1-cos^{2} \frac{\pi }{10}\right)\left(1-cos^{2} \frac{3\pi }{10}\right) = sin^{2} \frac{\pi }{10}. sin^{2} \frac{3\pi }{10}$
$= sin^{2}\,18^{\circ}. = sin^{2}\,54^{\circ} = \left(\frac{\sqrt{5}-1}{4}.\frac{\sqrt{5}+1}{4}\right)^{2} = \frac{1}{16}$