Q110 / 200 · Manipal (MET) 2011 · Mathematics
$\begin{aligned} &\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right) \\ &\left(1+\cos \frac{7 \pi}{8}\right)\end{aligned}$
is equal to
- A$\frac{1}{2}$
- Correct answerB$\frac{1}{8}$
- C$\cos \frac{\pi}{8}$
- D$\frac{1+\sqrt{2}}{2 \sqrt{2}}$
Solution
$\begin{aligned} & \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right) \left(1-\cos \frac{\pi}{8}\right) \\ & =\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right) \\ & =\left(\sin \frac{\pi}{8} \sin \frac{3 \pi}{8}\right)^2\end{aligned}$
$\begin{aligned} & =\frac{1}{4}\left(2 \sin \frac{3 \pi}{8} \sin \frac{\pi}{8}\right)^2 \\ & =\frac{1}{4}\left(\cos \frac{\pi}{4}-\cos \frac{\pi}{2}\right)^2=\frac{1}{8}\end{aligned}$
Hence, the correct answer is option: 'B'.