sum_{k=1}^3 cos^2 left((2k-1) frac{pi}{12}rig | vedclass

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(B) We need to evaluate the sum $S = \sum_{k=1}^3 \cos^2 \left((2k-1) \frac{\pi}{12}\right)$.Expanding the sum for $k=1, 2, 3$:$S = \cos^2 \left(\frac

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$\frac{3}{2}$

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(B) We need to evaluate the sum $S = \sum_{k=1}^3 \cos^2 \left((2k-1) \frac{\pi}{12}ight)$.Expanding the sum for $k=1, 2, 3$:$S = \cos^2 \left(\frac{\pi}{12}ight) + \cos^2 \left(\frac{3\pi}{12}ight) + \cos^2 \left(\frac{5\pi}{12}ight)$.Simplifying the angles:$S = \cos^2 \left(\frac{\pi}{12}ight) + \cos^2 \left(\frac{\pi}{4}ight) + \cos^2 \left(\frac{5\pi}{12}ight)$.Since $\cos \left(\frac{5\pi}{12}ight) = \cos \left(\frac{\pi}{2} - \frac{\pi}{12}ight) = \sin \left(\frac{\pi}{12}ight)$,we have $\cos^2 \left(\frac{5\pi}{12}ight) = \sin^2 \left(\frac{\pi}{12}ight)$.Substituting this into the sum:$S = \cos^2 \left(\frac{\pi}{12}ight) + \sin^2 \left(\frac{\pi}{12}ight) + \cos^2 \left(\frac{\pi}{4}ight)$.Using the identity $\cos^2 	heta + \sin^2 	heta = 1$:$S = 1 + \left(\frac{1}{\sqrt{2}}ight)^2 = 1 + \frac{1}{2} = \frac{3}{2}$.

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

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