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Question:
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$\begin{array}{r}
\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right) \\
\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{4 \pi}{8}\right) \\
\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{6 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=
\end{array}$
Options:
\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{7 \pi}{8}\right)=
\end{array}$
Solution:
2049 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{16}$
Given that,
$$
\begin{aligned}
& \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right) \\
& \left(1+\cos \frac{4 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{6 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right) \\
& =\left[1+\cos \frac{\pi}{8}\right]\left[1+\cos \frac{\pi}{4}\right]\left[1+\cos \frac{3 \pi}{8}\right]\left[1+\cos \frac{\pi}{2}\right] \\
& {\left[1+\cos \left(\pi-\frac{3 \pi}{8}\right)\right]\left[1+\cos \frac{3 \pi}{4}\right]\left[1+\cos \left(\pi-\frac{\pi}{8}\right)\right]}
\end{aligned}
$$
$\begin{aligned} & =\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{\pi}{4}\right)\left(1+\cos \frac{3 \pi}{8}\right) \\ & =\left(1+\cos \frac{\pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{4}\right)\left(1-\cos \frac{\pi}{8}\right) \\ & =\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right)\left(1-\frac{1}{2}\right) \\ & =\sin ^2 \frac{\pi}{8} \cdot \sin ^2 \frac{3 \pi}{8} \cdot \frac{1}{2}=\frac{1}{4}\left(2 \sin ^2 \frac{\pi}{8}\right)\left(2 \sin ^2 \frac{3 \pi}{8}\right) \cdot \frac{1}{2} \\ & =\frac{1}{4}\left(1-\cos \frac{\pi}{4}\right)\left(1-\cos \frac{3 \pi}{4}\right) \cdot \frac{1}{2} \\ & =\frac{1}{4} \cdot\left(1-\frac{1}{\sqrt{2}}\right)\left(1+\frac{1}{\sqrt{2}}\right) \cdot \frac{1}{2} \\ & =\frac{1}{4}\left(1-\frac{1}{2}\right) \cdot \frac{1}{2}=\frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{16} .\end{aligned}$
$$
\begin{aligned}
& \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right) \\
& \left(1+\cos \frac{4 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{6 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right) \\
& =\left[1+\cos \frac{\pi}{8}\right]\left[1+\cos \frac{\pi}{4}\right]\left[1+\cos \frac{3 \pi}{8}\right]\left[1+\cos \frac{\pi}{2}\right] \\
& {\left[1+\cos \left(\pi-\frac{3 \pi}{8}\right)\right]\left[1+\cos \frac{3 \pi}{4}\right]\left[1+\cos \left(\pi-\frac{\pi}{8}\right)\right]}
\end{aligned}
$$
$\begin{aligned} & =\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{\pi}{4}\right)\left(1+\cos \frac{3 \pi}{8}\right) \\ & =\left(1+\cos \frac{\pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{4}\right)\left(1-\cos \frac{\pi}{8}\right) \\ & =\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right)\left(1-\frac{1}{2}\right) \\ & =\sin ^2 \frac{\pi}{8} \cdot \sin ^2 \frac{3 \pi}{8} \cdot \frac{1}{2}=\frac{1}{4}\left(2 \sin ^2 \frac{\pi}{8}\right)\left(2 \sin ^2 \frac{3 \pi}{8}\right) \cdot \frac{1}{2} \\ & =\frac{1}{4}\left(1-\cos \frac{\pi}{4}\right)\left(1-\cos \frac{3 \pi}{4}\right) \cdot \frac{1}{2} \\ & =\frac{1}{4} \cdot\left(1-\frac{1}{\sqrt{2}}\right)\left(1+\frac{1}{\sqrt{2}}\right) \cdot \frac{1}{2} \\ & =\frac{1}{4}\left(1-\frac{1}{2}\right) \cdot \frac{1}{2}=\frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{16} .\end{aligned}$
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