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$\sum_{k=1}^3 \cos ^2\left((2 k-1) \frac{\pi}{12}\right)$ is equal to
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Verified Answer
The correct answer is:
$\frac{3}{4}$
We have
$$
\begin{aligned}
& \sum_{k=1}^3 \cos ^2\left[(2 k-1) \frac{\pi}{12}\right] \\
= & \cos ^2 \frac{\pi}{12}+\cos ^2 \frac{3 \pi}{12}+\cos ^2 \frac{5 \pi}{12} \\
= & \cos ^2 \frac{\pi}{12}+\frac{1}{2}+\cos ^2 \frac{5 \pi}{12} \\
= & \frac{1}{2}+\cos ^2 \frac{\pi}{12}+\cos ^2\left(\frac{\pi}{2}-\frac{\pi}{12}\right) \\
= & \frac{1}{2}+\cos ^2 \frac{\pi}{12}+\sin ^2 \frac{\pi}{12}=\frac{1}{2}+1=\frac{3}{2}
\end{aligned}
$$
$$
\begin{aligned}
& \sum_{k=1}^3 \cos ^2\left[(2 k-1) \frac{\pi}{12}\right] \\
= & \cos ^2 \frac{\pi}{12}+\cos ^2 \frac{3 \pi}{12}+\cos ^2 \frac{5 \pi}{12} \\
= & \cos ^2 \frac{\pi}{12}+\frac{1}{2}+\cos ^2 \frac{5 \pi}{12} \\
= & \frac{1}{2}+\cos ^2 \frac{\pi}{12}+\cos ^2\left(\frac{\pi}{2}-\frac{\pi}{12}\right) \\
= & \frac{1}{2}+\cos ^2 \frac{\pi}{12}+\sin ^2 \frac{\pi}{12}=\frac{1}{2}+1=\frac{3}{2}
\end{aligned}
$$
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