$\sum_{k=1}^3 \cos ^2\left((2 k-1) \frac{\pi}{12}\right)$ is equal to

Question

Aicharya · Accepted Answer

We have $ \begin{aligned} & \sum_{k=1}^3 \cos ^2\left[(2 k-1) \frac{\pi}{12}ight] \ = & \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}ight) \ = & \frac{1}{2}+\cos ^2 \frac{\pi}{12}+\sin ^2 \frac{\pi}{12}=\frac{1}{2}+1=\frac{3}{2} \end{aligned} $

$\sum_{k=1}^3 \cos ^2\left((2 k-1) \frac{\pi}{12}\right)$ is equal to

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