Fun with 2018 #2

Geometry Level 3

sin ⁡ ( π 1009 ) + sin ⁡ ( 2 π 1009 ) + sin ⁡ ( 3 π 1009 ) + … + sin ⁡ ( 1008 π 1009 ) = cos ⁡ ( π 2018 ) − cos ⁡ ( A π 2018 ) 2 sin ⁡ ( π 2018 ) \sin \left(\frac{\pi}{1009}\right) + \sin \left(\frac{2\pi}{1009}\right) + \sin \left(\frac{3\pi}{1009}\right) + \ldots + \sin \left(\frac{1008\pi}{1009}\right) = \frac{\cos \left(\frac{\pi}{2018}\right) - \cos \left(\frac{A \pi}{2018}\right)}{2\sin \left(\frac{\pi}{2018}\right)}

What is the smallest positive integer A A such that the equation above is satisfied?


The answer is 2017.

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Guillermo Templado by Guillermo Templado , 46, Spain

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1 solution

Relevant wiki: Euler's Formula

S = ∑ k = 1 1008 sin ⁡ ( k π 1009 ) By Euler’s formula: e θ i = cos ⁡ θ + i sin ⁡ θ = ℑ ( ∑ k = 1 1008 e k π 1009 i ) where ℑ ( z ) is the imaginary part of z . = ℑ ( e π 1009 i ( 1 − e 1008 π 1009 i 1 − e π 1009 i ) ) = ℑ ( e π 1009 i − e π i 1 − e π 1009 i ) \begin{aligned} S & = \sum_{k=1}^{1008} \sin \left(\frac {k\pi}{1009}\right) & \small \color{#3D99F6} \text{By Euler's formula: } e^{\theta i} = \cos \theta + i \sin \theta \\ & = \Im \left(\sum_{k=1}^{1008} e^{\frac {k\pi}{1009}i} \right) & \small \color{#3D99F6} \text{where }\Im(z) \text{ is the imaginary part of }z. \\ & = \Im \left(e^{\frac {\pi}{1009}i}\left(\frac {1-e^{\frac {1008\pi}{1009}i}}{1-e^{\frac {\pi}{1009}i}} \right)\right) \\ & = \Im \left(\frac {e^{\frac {\pi}{1009}i}-e^{\pi i}}{1-e^{\frac {\pi}{1009}i}}\right) \end{aligned}

= ℑ ( cos ⁡ ( π 1009 ) + i sin ⁡ ( π 1009 ) + 1 1 − cos ⁡ ( π 1009 ) − i sin ⁡ ( π 1009 ) ) = ℑ ( ( 1 + cos ⁡ ( π 1009 ) + i sin ⁡ ( π 1009 ) ) ( 1 − cos ⁡ ( π 1009 ) + i sin ⁡ ( π 1009 ) ) ( 1 − cos ⁡ ( π 1009 ) − i sin ⁡ ( π 1009 ) ) ( 1 − cos ⁡ ( π 1009 ) + i sin ⁡ ( π 1009 ) ) ) = ℑ ( 2 i sin ⁡ ( π 1009 ) 2 − 2 cos ⁡ ( π 1009 ) ) = sin ⁡ ( π 1009 ) 1 − cos ⁡ ( π 1009 ) = 2 sin ⁡ ( π 2018 ) cos ⁡ ( π 2018 ) 1 − ( 1 − 2 sin ⁡ 2 ( π 2018 ) ) = 2 cos ⁡ ( π 2018 ) 2 sin ⁡ ( π 2018 ) \begin{aligned} \ \ & = \Im \left(\frac {\cos \left(\frac {\pi}{1009} \right) + i\sin \left(\frac {\pi}{1009} \right) + 1}{1-\cos \left(\frac {\pi}{1009} \right) - i\sin \left(\frac {\pi}{1009} \right) }\right) \\ & = \Im \left(\frac {\left(1 + \cos \left(\frac {\pi}{1009} \right) + i\sin \left(\frac {\pi}{1009} \right) \right) \left(1-\cos \left(\frac {\pi}{1009} \right) + i\sin \left(\frac {\pi}{1009} \right) \right) }{\left(1-\cos \left(\frac {\pi}{1009} \right) - i\sin \left(\frac {\pi}{1009} \right) \right) \left(1-\cos \left(\frac {\pi}{1009} \right) + i\sin \left(\frac {\pi}{1009} \right) \right)}\right) \\ & = \Im \left(\frac {2 i \sin \left(\frac {\pi}{1009} \right)}{2-2\cos \left(\frac {\pi}{1009} \right)}\right) \\ & = \frac {\sin \left(\frac {\pi}{1009} \right)}{1-\cos \left(\frac {\pi}{1009} \right)} \\ & = \frac {2\sin \left(\frac {\pi}{2018} \right)\cos \left(\frac {\pi}{2018} \right)}{1-\left(1 - 2\sin^2 \left(\frac {\pi}{2018} \right)\right)} \\ & = \frac {2 \cos \left(\frac {\pi}{2018} \right)}{2\sin \left(\frac {\pi}{2018} \right)} \end{aligned}

= cos ⁡ ( π 2018 ) + cos ⁡ ( π 2018 ) 2 sin ⁡ ( π 2018 ) Note that cos ⁡ ( π − θ ) = − cos ⁡ θ = cos ⁡ ( π 2018 ) − cos ⁡ ( 2017 π 2018 ) 2 sin ⁡ ( π 2018 ) \begin{aligned} \ \ & = \frac {\cos \left(\frac {\pi}{2018} \right) + \color{#3D99F6} \cos \left(\frac {\pi}{2018} \right)}{2\sin \left(\frac {\pi}{2018} \right)} & \small \color{#3D99F6} \text{Note that }\cos (\pi - \theta) = - \cos \theta \\ & = \frac {\cos \left(\frac {\pi}{2018} \right) - \color{#3D99F6} \cos \left(\frac {2017\pi}{2018} \right)}{2\sin \left(\frac {\pi}{2018} \right)} \end{aligned}

Therefore, A = 2017 A = \boxed{2017} .

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