Q60 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics
$\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)$
$$
\left(1+\cos \frac{7 \pi}{8}\right)=
$$
- A$\frac{1+\sqrt{2}}{2 \sqrt{2}}$
- B$\frac{\pi}{8}$
- Correct answerC$\frac{1}{8}$
- D$\frac{1}{2}$
Solution
We have,
$$
\begin{gathered}
\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right) \\
\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right) \\
=\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right) \\
=\sin ^2 \frac{\pi}{8} \sin ^2 \frac{3 \pi}{8}=\sin ^2 \frac{\pi}{8} \sin ^2\left(\frac{\pi}{2}-\frac{\pi}{8}\right) \\
=\sin ^2 \frac{\pi}{8} \cos ^2 \frac{\pi}{8}=\frac{1}{4}\left(2 \sin \frac{\pi}{8} \cos \frac{\pi}{8}\right)^2 \\
=\frac{1}{4} \sin ^2 \frac{\pi}{4}=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}
\end{gathered}
$$
Hence, the correct answer is option: 'C'.