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