A geometry problem by A Former Brilliant Member

Geometry Level 4

cos ⁡ 2 ( π n ) + cos ⁡ 2 ( 2 π n ) + cos ⁡ 2 ( 3 π n ) + ⋯ + cos ⁡ 2 ( ( n − 1 ) π n ) \cos^2 \left(\dfrac\pi n \right) + \cos^2 \left( \dfrac {2\pi} n \right) + \cos^2 \left(\dfrac {3\pi} n \right) + \cdots + \cos^2 \left(\dfrac {(n-1)\pi} n \right)

For integer n > 1 n>1 , find the closed form of the expression above.

n − 1 2 \dfrac{n-1}{2} n ( n + 1 ) 2 \dfrac{n(n+1)}{2} n 2 \dfrac{n}{2} n − 2 2 \dfrac{n-2}{2} None of these choices

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A Former Brilliant Member by A Former Brilliant Member

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1 solution

S = ∑ k = 1 n − 1 cos ⁡ 2 k π n = 1 2 ∑ k = 1 n − 1 ( 1 + cos ⁡ 2 k π n ) = n − 1 2 + 1 2 ∑ k = 1 n − 1 cos ⁡ 2 k π n \begin{aligned} S & = \sum_{k=1}^{n-1} \cos^2 \frac {k\pi}n = \frac 12 \sum_{k=1}^{n-1} \left(1+\cos \frac {2k\pi}n \right) = \frac {n-1}2 + \frac 12 \sum_{k=1}^{n-1} \cos \frac {2k\pi}n \end{aligned}

For even n = 2 m n=2m , then

∑ k = 1 n − 1 cos ⁡ 2 k π n = ∑ k = 1 2 m − 1 cos ⁡ k π m = ∑ k = 1 m − 1 cos ⁡ k π m + cos ⁡ m π m + ∑ k = m + 1 2 m − 1 cos ⁡ k π m = ∑ k = 1 m − 1 cos ⁡ k π m − 1 + ∑ k = 1 m − 1 cos ⁡ ( m + k ) π m = ∑ k = 1 m − 1 cos ⁡ k π m − 1 + ∑ k = 1 m − 1 cos ⁡ ( π + k π m ) Note that cos ⁡ ( π + θ ) = − cos ⁡ θ = ∑ k = 1 m − 1 cos ⁡ k π m − 1 − ∑ k = 1 m − 1 cos ⁡ k π m = − 1 \begin{aligned} \sum_{k=1}^{n-1} \cos \frac {2k\pi}n & = \sum_{k=1}^{2m-1} \cos \frac {k\pi}m \\ & = \sum_{k=1}^{m-1} \cos \frac {k\pi}m + \cos \frac {m\pi}m + \sum_{k=m+1}^{2m-1} \cos \frac {k\pi}m \\ & = \sum_{k=1}^{m-1} \cos \frac {k\pi}m - 1 + \sum_{k=1}^{m-1} \cos \frac {(m+k)\pi}m \\ & = \sum_{k=1}^{m-1} \cos \frac {k\pi}m - 1 + \sum_{k=1}^{m-1} \cos \left(\pi + \frac {k\pi}m\right) & \small \color{#3D99F6} \text{Note that }\cos (\pi + \theta) = - \cos \theta \\ & = \sum_{k=1}^{m-1} \cos \frac {k\pi}m - 1 - \sum_{k=1}^{m-1} \cos \frac {k\pi}m \\ & = - 1 \end{aligned}

For odd n = 2 m + 1 n=2m+1 , then

∑ k = 1 2 m cos ⁡ 2 k π 2 m + 1 = ∑ k = 1 m cos ⁡ 2 k π 2 m + 1 + ∑ k = m + 1 2 m cos ⁡ 2 k π 2 m + 1 Note that ∑ k = 1 m cos ⁡ 2 k π 2 m + 1 = − 1 2 = − 1 2 + ∑ k = 1 m cos ⁡ 2 ( m + k ) π 2 m + 1 = − 1 2 + ∑ k = 1 m cos ⁡ ( π − ( 2 k − 1 ) π 2 m + 1 ) = − 1 2 − ∑ k = 1 m cos ⁡ ( 2 k − 1 ) π 2 m + 1 and ∑ k = 1 m cos ⁡ ( 2 k − 1 ) π 2 m + 1 = 1 2 = − 1 2 − 1 2 = − 1 \begin{aligned} \sum_{k=1}^{2m} \cos \frac {2k\pi}{2m+1} & = {\color{#3D99F6}\sum_{k=1}^m \cos \frac {2k\pi}{2m+1}} + \sum_{k=m+1}^{2m} \cos \frac {2k\pi}{2m+1} & \small \color{#3D99F6} \text{Note that } \sum_{k=1}^m \cos \frac {2k\pi}{2m+1} = - \frac 12 \\ & = {\color{#3D99F6} - \frac 12} + \sum_{k=1}^m \cos \frac {2(m+k)\pi}{2m+1} \\ & = - \frac 12 + \sum_{k=1}^m \cos \left(\pi - \frac {(2k-1)\pi}{2m+1} \right) \\ & = - \frac 12 - \color{#3D99F6} \sum_{k=1}^m \cos \frac {(2k-1)\pi}{2m+1} & \small \color{#3D99F6} \text{and } \sum_{k=1}^m \cos \frac {(2k-1)\pi}{2m+1} = \frac 12 \\ & = - \frac 12 - \color{#3D99F6} \frac 12 \\ & = - 1 \end{aligned}

Therefore, for all n > 1 n > 1 , S = ∑ k = 1 n − 1 cos ⁡ 2 k π n = n − 1 2 + 1 2 ( − 1 ) = n − 2 2 \displaystyle S = \sum_{k=1}^{n-1} \cos^2 \frac {k\pi}n = \frac {n-1}2 + \frac 12 (-1) = \boxed{\dfrac {n-2}2} .

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