Resolved: "Why is:∑k=1[n2]sin2((2k−1)πn)=n4"

Scoopedepalj

Answered question

2022-02-28

Why is:

\sum_{k = 1}^{[\frac{n}{2}]} \sin^{2} ((2 k - 1) \frac{π}{n}) = \frac{n}{4}

chezpepina87j

Beginner2022-03-01Added 6 answers

Let

T = \sum_{k = 1}^{[\frac{n}{2}]} \cos ((2 k - 1) \frac{2 π}{n})

S = \sum_{k = 1}^{[\frac{n}{2}]} \sin^{2} ((2 k - 1) \frac{π}{n})

We can get the relation

T + 2 S = [\frac{n}{2}]

If n is even then using summation of cosines whose angles are in AP we get,

T = \sum_{k = 1}^{[\frac{n}{2}]} \cos ((2 k - 1) \frac{2 π}{n}) = \frac{\sin [\frac{n}{2}] \frac{2 π}{n}}{\sin \frac{2 π}{n}} \cos [\frac{2 π}{n} + ([\frac{n}{2}] - 1) \frac{2 π}{n}] = 0

\Rightarrow S = \frac{n}{4}

If n is odd then let

n = 2 m + 1

. We get,

T = \sum_{k = 1}^{m} \cos ((2 k - 1) \frac{2 π}{2 m + 1})

= \frac{\sin \frac{2 π m}{2 m + 1}}{\sin \frac{2 π}{2 m + 1}} \cos [\frac{2 π}{2 m + 1} + (m - 1) \frac{2 π}{2 m + 1}]

= \frac{\sin \frac{2 π m}{2 m + 1} \cos \frac{2 π m}{2 m + 1}}{\sin \frac{2 π}{2 m + 1}}

= \frac{- \sin \frac{2 π}{2 m + 1}}{2 \sin \frac{2 π}{2 m + 1}} = \frac{- 1}{2}

\Rightarrow S = \frac{2 m + 1}{2} = \frac{n}{4}

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