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Question: Answered & Verified by Expert
$$
\begin{aligned}
& \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{4 \pi}{8}\right) \ldots \\
& \left(1+\cos \frac{7 \pi}{8}\right)=
\end{aligned}
$$
MathematicsTrigonometric EquationsAP EAMCETAP EAMCET 2023 (18 May Shift 2)
Options:
  • A $\frac{1}{16}$
  • B $\frac{1}{64}$
  • C $\frac{3}{16}$
  • D $\frac{3}{64}$
Solution:
1223 Upvotes Verified Answer
The correct answer is: $\frac{1}{16}$
$\begin{aligned} & \text { }\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{4 \pi}{8}\right) \\ & \left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{6 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right) \\ & =\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{\pi}{4}\right)\left(1+\cos \frac{3 \pi}{4}\right)\left(1+\cos \frac{\pi}{2}\right) \\ & \left(1+\cos \left(\pi-\frac{3 \pi}{8}\right)\right)\left(1+\cos \frac{3 \pi}{4}\right)\left(1+\cos \left(\pi-\frac{\pi}{8}\right)\right) \\ & =\left(1+\cos \frac{\pi}{8}\right)\left(1+\frac{1}{\sqrt{2}}\right)\left(1+\cos \frac{3 \pi}{8}\right)(1+0) \\ & \left(1-\cos \frac{3 \pi}{8}\right)\left(1-\frac{1}{\sqrt{2}}\right)\left(1-\cos \frac{\pi}{8}\right) \\ & =\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right)\left(1-\frac{1}{2}\right) \\ & =\sin ^2 \frac{\pi}{8} \sin ^2 \frac{3 \pi}{8} \frac{1}{2} \\ & =\sin ^2 \frac{\pi}{8} \cos ^2\left(\frac{\pi}{2}-\frac{n}{8}\right) \times \frac{1}{2} \\ & =\sin ^2 \frac{\pi}{8} \cos ^2 \frac{\pi}{8} \times \frac{1}{2} \\ & =\frac{1}{4}\left(2 \sin \frac{\pi}{8} \cos \frac{\pi}{8}\right)^2 \times \frac{1}{2} \\ & =\frac{1}{8}\left(\sin \frac{\pi}{4}\right)^2=\frac{1}{8} \times\left(\frac{1}{\sqrt{2}}\right)^2=\frac{1}{16} \\ & \end{aligned}$

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