Question

$\left(1+\cos \frac{\pi }{8}\right)\left(1+\cos \frac{3\pi }{8}\right)\left(1+\cos \frac{5\pi }{8}\right)$
$\left(1+\cos \frac{7\pi }{8}\right)=$

• A. $\frac{1+\sqrt{2}}{2\sqrt{2}}$
• B. $\frac{\pi }{8}$
• C. $\frac{1}{8}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

We have,
$\left(1+\cos \frac{\pi }{8}\right)\left(1+\cos \frac{3\pi }{8}\right)\left(1+\cos \frac{5\pi }{8}\right)\left(1+\cos \frac{7\pi }{8}\right)$
$\left(1+\cos \frac{\pi }{8}\right)\left(1+\cos \frac{3\pi }{8}\right)\left(1-\cos \frac{3\pi }{8}\right)\left(1-\cos \frac{\pi }{8}\right)$
$[\because \cos (\pi -\theta )=\cos \theta ]$
$=\left(1-{\cos }^2\frac{\pi }{8}\right)\left(1-{\cos }^2\frac{3\pi }{8}\right)$
$={\sin }^2\frac{\pi }{8}{\sin }^2\frac{3\pi }{8}={\sin }^2\frac{\pi }{8}{\sin }^2\left(\frac{\pi }{2}-\frac{\pi }{8}\right)$
$={\sin }^2\frac{\pi }{8}{\cos }^2\frac{\pi }{8}=\frac{1}{4}{\left(2\sin \frac{\pi }{8}\cos \frac{\pi }{8}\right)}^2$
$=\frac{1}{4}{\sin }^2\frac{\pi }{4}=\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}$