∑k=13cos2(2k-1)π12 =

∑k=13 cos2 (2k-1) π12 =cos2π12+cos2π4+cos25π12 A+B=π2 ⇒ cos2A+cos2B=1 =1+12=32

∑k=13cos2(2k-1)π12 =

∑k=13 cos2 (2k-1) π12 =cos2π12+cos2π4+cos25π12 A+B=π2 ⇒ cos2A+cos2B=1 =1+12=32

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$\sum_{k = 1}^{3} \cos^{2} (2 k - 1) \frac{π}{12} =$

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$\frac{- 1}{2}$

$\frac{1}{2}$

$\frac{3}{2}$

answer is D.

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Detailed Solution

$\begin{array}{rcl} \sum_{k = 1}^{3} \cos^{2} (2 k - 1) \frac{π}{12} \\ = & \cos^{2} \frac{π}{12} + \cos^{2} \frac{π}{4} + \cos^{2} \frac{5 π}{12} \\ \begin{array}{rcl} A + B & = & \frac{π}{2} \Rightarrow \cos^{2} A + \cos^{2} B = 1 \end{array} \\ = & 1 + \frac{1}{2} = \frac{3}{2} \end{array}$