Damned if you do it by trigo.

Calculus Level 5

Evaluate the below expression:

∑ k = 1 n ( 2 k − 1 ) sin ⁡ ( 2 k − 1 ) x \displaystyle \sum_{k=1}^n (2k-1)\sin(2k-1)x

1 4 sin ⁡ 2 x [ ( 2 n − 1 ) sin ⁡ ( 2 n − 1 ) x − ( 2 n + 1 ) sin ⁡ ( 2 n + 1 ) x ] \dfrac{1}{4\sin^2x}[(2n-1)\sin(2n-1)x - (2n+1)\sin(2n+1)x] 1 4 sin ⁡ 2 x [ ( 2 n + 1 ) sin ⁡ ( 2 n + 1 ) x − ( 2 n − 1 ) sin ⁡ ( 2 n − 1 ) x ] \dfrac{1}{4\sin^2x}[(2n+1)\sin(2n+1)x - (2n-1)\sin(2n-1)x] 1 4 sin ⁡ 2 x [ ( 2 n − 1 ) sin ⁡ ( 2 n + 1 ) x − ( 2 n + 1 ) sin ⁡ ( 2 n − 1 ) x ] \dfrac{1}{4\sin^2x}[(2n-1)\sin(2n+1)x - (2n+1)\sin(2n-1)x] − 1 2 sin ⁡ 2 x [ ( 2 n + 1 ) sin ⁡ ( 2 n − 1 ) x − ( 2 n − 1 ) sin ⁡ ( 2 n + 1 ) x ] \dfrac{-1}{2\sin^2x}[(2n+1)\sin(2n-1)x - (2n-1)\sin(2n+1)x] 1 4 sin ⁡ 2 x [ ( 2 n + 1 ) sin ⁡ ( 2 n − 1 ) x − ( 2 n − 1 ) sin ⁡ ( 2 n + 1 ) x ] \dfrac{1}{4\sin^2x}[(2n+1)\sin(2n-1)x - (2n-1)\sin(2n+1)x]

Vishwak Srinivasan by Vishwak Srinivasan , 24, India

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1 solution

Let us consider the sum S S , which is

S = ∑ k = 1 n cos ⁡ ( 2 k − 1 ) x S = \displaystyle \sum_{k=1}^{n} \cos(2k-1)x

S = cos ⁡ x + cos ⁡ ( 3 ) x + cos ⁡ ( 5 ) x + . . . . . . . . + cos ⁡ ( 2 n − 1 ) x S = \cos x + \cos(3)x + \cos(5)x + . . . . . . . . + \cos(2n-1)x

Differentiating on both sides,

S 1 = S ′ = − ( sin ⁡ x + 3 sin ⁡ ( 3 ) x + 5 sin ⁡ ( 5 ) x + . . . . . . + ( 2 n − 1 ) sin ⁡ ( 2 n − 1 ) x ) S_{1} = S' = -(\sin x + 3\sin(3)x + 5\sin(5)x + . . . . . . + (2n-1)\sin(2n-1)x)

S S can be found using the "angles in an A.P formula".

Differentiate S S to get S ′ = S 1 = − 1 × r e q d . s u m S' = S_{1} = -1 \times reqd. sum

Rest is trivial.

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