∑ _k=1^3cos^2 [(2k 1)π2] =cos^2π12+cos^23π12+cos^25π12 cos^2π12+sin^2π12+cos^2π4 [∴ cos (π2 θ) =sinθ⇒ cos (π2 5π2)=sinπ12] =1+cos^2π4 ...

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Integration by Substitution

∑k=13cos2[2k-...

Question

$3 \sum k = 1 c o s^{2} [(2 k - 1) \frac{π}{12}] =$

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Solution

$3 \sum k = 1 {cos}^{2} [(2 k - 1) \frac{π}{2}]$
$= {cos}^{2} \frac{π}{12} + {cos}^{2} \frac{3 π}{12} + {cos}^{2} \frac{5 π}{12}$
${cos}^{2} \frac{π}{12} + {sin}^{2} \frac{π}{12} + {cos}^{2} \frac{π}{4} [∴ cos (\frac{π}{2} - θ) = sin θ \Rightarrow cos (\frac{π}{2} - \frac{5 π}{2}) = sin \frac{π}{12}]$
$= 1 + {cos}^{2} \frac{π}{4}$
$= 1 + \frac{1}{2} = \frac{3}{2}$

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