# The value of

Question:

The value of $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$ is ___________________

Solution:

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

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

$=\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right) \quad[\because \cos (\pi-\theta=-\cos \theta)]$

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

$=\left(1-\cos ^{2} \frac{\pi}{8}\right)\left(1-\cos ^{2} 3 \frac{\pi}{8}\right)$

$=\sin ^{2} \pi / 8 \sin ^{2} 3 \pi / 8$

$=\sin ^{2} \pi / 8 \sin ^{2}\left(\frac{\pi}{2}-\frac{\pi}{2}\right)$

$=\sin ^{2} \pi / 8 \cos ^{2} \pi / 8 \quad(\because \sin (\pi / 2)=\cos \theta)$

$=\frac{1}{4}(2 \sin \pi / 8 \cos \pi / 8)^{2}$

$=\frac{1}{4}\left(\sin 2 \times \frac{\pi}{8}\right)^{2}=\frac{1}{4}\left(\sin \frac{\pi}{4}\right)^{2}$

$=\frac{1}{4}\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}$

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