The value of sum_{k=1}^3 cos ^2left((2 k-1) f | vedclass

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(C) We have $\sum_{k=1}^3 \cos ^2\left((2 k-1) \frac{\pi}{12}\right) = \cos ^2 \frac{\pi}{12} + \cos ^2 \frac{3 \pi}{12} + \cos ^2 \frac{5 \pi}{12}$.S

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) We have $\sum_{k=1}^3 \cos ^2\left((2 k-1) \frac{\pi}{12}ight) = \cos ^2 \frac{\pi}{12} + \cos ^2 \frac{3 \pi}{12} + \cos ^2 \frac{5 \pi}{12}$.Since $\frac{3 \pi}{12} = \frac{\pi}{4}$,we have $\cos ^2 \frac{\pi}{4} = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.Thus,the expression becomes $\cos ^2 \frac{\pi}{12} + \frac{1}{2} + \cos ^2 \frac{5 \pi}{12}$.Using the identity $\cos(\frac{\pi}{2} - 	heta) = \sin 	heta$,we have $\cos \frac{5 \pi}{12} = \cos(\frac{\pi}{2} - \frac{\pi}{12}) = \sin \frac{\pi}{12}$.Substituting this,we get $\frac{1}{2} + \cos ^2 \frac{\pi}{12} + \sin ^2 \frac{\pi}{12}$.Since $\cos ^2 	heta + \sin ^2 	heta = 1$,the expression simplifies to $\frac{1}{2} + 1 = \frac{3}{2}$.

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

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