Play with sin

Geometry Level 2

2 sin ⁡ 2 ∘ + 4 sin ⁡ 4 ∘ + 6 sin ⁡ 6 ∘ + ⋯ + 180 sin ⁡ 18 0 ∘ = ? 2 \sin 2^\circ + 4 \sin 4^\circ + 6 \sin 6^\circ + \cdots + 180 \sin 180^\circ =\ ?

90 90 90 cos ⁡ 1 ∘ 90 \cos 1^\circ 90 cot ⁡ 1 ∘ 90 \cot 1^\circ 90 sin ⁡ 1 ∘ 90 \sin 1^\circ

Aly Ahmed by Aly Ahmed , 34, Egypt

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1 solution

Chew-Seong Cheong
May 13, 2020

The given sum can be written as:

S = ∑ k = 1 90 2 k sin ⁡ ( 2 k ∘ ) = ∑ k = 1 89 2 k sin ⁡ ( 2 k ∘ ) Since sin ⁡ 18 0 ∘ = 0 = ∑ k = 1 44 2 k sin ⁡ ( 2 k ∘ ) + 90 sin ⁡ 9 0 ∘ + ∑ k = 46 89 2 k sin ⁡ ( 2 k ∘ ) = ∑ k = 1 44 2 k sin ⁡ ( 2 k ∘ ) + 90 + ∑ k = 1 44 ( 180 − 2 k ) sin ⁡ ( 18 0 ∘ − 2 k ∘ ) Note that sin ⁡ ( 18 0 ∘ − θ ) = sin ⁡ θ = 180 ∑ k = 1 44 sin ⁡ ( 2 k ∘ ) + 90 = 180 ( ∑ k = 1 22 sin ⁡ ( 2 k ∘ ) + ∑ k = 23 44 sin ⁡ ( 2 k ∘ ) ) + 90 = 180 ( ∑ k = 1 22 sin ⁡ ( 2 k ∘ ) + ∑ k = 23 44 cos ⁡ ( 9 0 ∘ − 2 k ∘ ) ) + 90 = 180 ( ∑ k = 1 22 sin ⁡ ( 2 k ∘ ) + ∑ k = 1 22 cos ⁡ ( 2 k ∘ ) ) + 90 By Euler’s formula: e θ i = cos ⁡ θ + i sin ⁡ θ = 180 ( ℑ ( ∑ k = 1 22 e 2 k ∘ i ) + ℜ ( ∑ k = 1 22 e 2 k ∘ i ) ) + 90 where ℑ ( ⋅ ) is the imaginary part function = 180 ( ℑ ( e 2 ∘ i ( 1 − e 4 4 ∘ i ) 1 − e 2 ∘ i ) + ℜ ( ∑ k = 1 22 e 2 k ∘ i ) ) + 90 and ℜ ( ⋅ ) , the real part function. = 180 ( ℑ ( e 2 4 ∘ i ( e − 2 2 ∘ i − e 2 2 ∘ i ) e 1 ∘ i ( e − 1 ∘ i − e 1 ∘ i ) + ℜ ( ∑ k = 1 22 e 2 k ∘ i ) ) + 90 = 180 ( ℑ ( ( cos ⁡ 2 3 ∘ + i sin ⁡ 2 3 ∘ ) sin ⁡ 2 2 ∘ sin ⁡ 1 ∘ ) + ℜ ( ∑ k = 1 22 e 2 k ∘ i ) ) + 90 = 180 ( ℑ ( cos ⁡ 2 3 ∘ cos ⁡ 6 8 ∘ + i sin ⁡ 2 3 ∘ sin ⁡ 2 2 ∘ sin ⁡ 1 ∘ ) + ℜ ( ∑ k = 1 22 e 2 k ∘ i ) ) + 90 = 180 ( ℑ ( cos ⁡ 4 5 ∘ + cos ⁡ 9 1 ∘ + i ( cos ⁡ 1 ∘ − cos ⁡ 4 5 ∘ ) 2 sin ⁡ 1 ∘ ) + ℜ ( ∑ k = 1 22 e 2 k ∘ i ) ) + 90 = 90 ( cos ⁡ 4 5 ∘ − sin ⁡ 1 ∘ + cos ⁡ 1 ∘ − cos ⁡ 4 5 ∘ sin ⁡ 1 ∘ ) + 90 = 90 cot ⁡ 1 ∘ \begin{aligned} S & = \sum_{k=1}^\blue{90} 2k \sin (2k^\circ) = \sum_{k=1}^\red{89} 2k \sin (2k^\circ) & \small \blue{\text{Since }\sin 180^\circ=0} \\ & = \sum_{k=1}^{44} 2k \sin (2k^\circ) + 90 \sin 90^\circ + \sum_{k=46}^{89} 2k \sin (2k^\circ) \\ & = \sum_{k=1}^{44} 2k \sin (2k^\circ) + 90 + \sum_{k=1}^{44} (180-2k) \blue{\sin (180^\circ - 2k^\circ)} & \small \blue{\text{Note that }\sin (180^\circ - \theta) = \sin \theta} \\ & = 180 \sum_{k=1}^{44} \sin (2k^\circ) + 90 \\ & = 180 \left(\sum_{k=1}^{22} \sin (2k^\circ) + \sum_{k=23}^{44} \sin (2k^\circ) \right) + 90 \\ & = 180 \left(\sum_{k=1}^{22} \sin (2k^\circ) + \sum_{k=23}^{44} \cos (90^\circ - 2k^\circ) \right) + 90 \\ & = 180 \left(\sum_{k=1}^{22} \sin (2k^\circ) + \sum_{k=1}^{22} \cos (2k^\circ) \right) + 90 & \small \blue{\text{By Euler's formula: }e^{\theta i} = \cos \theta + i \sin \theta} \\ & = 180 \left(\Im \left(\sum_{k=1}^{22} e^{2k^\circ i} \right) + \Re \left(\sum_{k=1}^{22} e^{2k^\circ i} \right)\right) + 90 & \small \blue{\text{where } \Im (\cdot) \text{ is the imaginary part function}} \\ & = 180 \left(\Im \left(\frac {e^{2^\circ i}(1-e^{44^\circ i})}{1-e^{2^\circ i}} \right) + \Re \left(\sum_{k=1}^{22} e^{2k^\circ i} \right)\right) + 90 & \small \blue{\text{and } \Re (\cdot) \text{, the real part function.}} \\ & = 180 \left(\Im \left(\frac {e^{24^\circ i}(e^{-22^\circ i}-e^{22^\circ i})}{e^{1^\circ i}(e^{-1^\circ i}-e^{1^\circ i}} \right) + \Re \left(\sum_{k=1}^{22} e^{2k^\circ i} \right)\right) + 90 \\ & = 180 \left(\Im \left(\frac {(\cos 23^\circ + i\sin 23^\circ)\sin 22^\circ}{\sin 1^\circ} \right) + \Re \left(\sum_{k=1}^{22} e^{2k^\circ i} \right)\right) + 90 \\ & = 180 \left(\Im \left(\frac {\cos 23^\circ \cos 68^\circ + i\sin 23^\circ\sin 22^\circ}{\sin 1^\circ} \right) + \Re \left(\sum_{k=1}^{22} e^{2k^\circ i} \right)\right) + 90 \\ & = 180 \left(\Im \left(\frac {\cos 45^\circ + \cos 91^\circ + i(\cos 1^\circ - \cos 45^\circ)}{2 \sin 1^\circ} \right) + \Re \left(\sum_{k=1}^{22} e^{2k^\circ i} \right)\right) + 90 \\ & = 90 \left(\frac {\cos 45^\circ - \sin 1^\circ + \cos 1^\circ - \cos 45^\circ}{\sin 1^\circ} \right) + 90 \\ & = \boxed{90\cot 1^\circ} \end{aligned}

Reference: Euler's formula

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