100 Followers Problem

Geometry Level 4

∑ k = 1 35 sin ⁡ ( 5 k ∘ ) = tan ⁡ ( m n ) ∘ \large \sum_{k=1}^{35} \sin \left(5k^\circ \right) = \tan \left(\frac {m}{n}\right)^\circ .

The equation above holds true for coprime positive integers m m and n n , where m n < 90 \dfrac {m}{n} < 90 . Find m + n m+n .

173 175 179 177

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Md Zuhair by Md Zuhair , 20, India

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2 solutions

Chew-Seong Cheong
May 16, 2017

S = ∑ k = 1 35 sin ⁡ ( 5 k ∘ ) By Euler’s formula: e x i = cos ⁡ x + i sin ⁡ x = ∑ k = 1 35 ℑ [ e 5 k ∘ i ] where ℑ ( z ) is the imaginary part of z . = ℑ [ ∑ k = 1 35 e 5 k ∘ i ] Summation of geometric progression = ℑ [ e 5 ∘ i ( 1 − e 17 5 ∘ i 1 − e 5 ∘ i ) ] = ℑ [ e 5 ∘ i − e 18 0 ∘ i 1 − e 5 ∘ i ] = ℑ [ e 5 ∘ i + 1 1 − e 5 ∘ i ] = ℑ [ 1 + cos ⁡ 5 ∘ + i sin ⁡ 5 ∘ 1 − cos ⁡ 5 ∘ − i sin ⁡ 5 ∘ ] Multiply up and down with 1 − cos ⁡ 5 ∘ + i sin ⁡ 5 ∘ = ℑ [ 2 i sin ⁡ 5 ∘ 2 − 2 cos ⁡ 5 ∘ ] = sin ⁡ 5 ∘ 1 − cos ⁡ 5 ∘ = 2 sin ⁡ 2. 5 ∘ cos ⁡ 2. 5 ∘ 1 − 1 + 2 sin ⁡ 2 2. 5 ∘ = cot ⁡ 5 ∘ 2 = tan ⁡ 17 5 ∘ 2 \begin{aligned} S & = \sum_{k=1}^{35} \sin \left(5k^\circ \right) & \small \color{#3D99F6} \text{By Euler's formula: }e^{xi} = \cos x + i \sin x \\ & = \sum_{k=1}^{35} \Im \left[e^{5k^\circ i}\right] & \small \color{#3D99F6} \text{where } \Im(z) \text{ is the imaginary part of }z. \\ & = \Im \left[\sum_{k=1}^{35} e^{5k^\circ i}\right] & \small \color{#3D99F6} \text{Summation of geometric progression} \\ & = \Im \left[e^{5^\circ i} \left( \frac {1-e^{175^\circ i}}{1-e^{5^\circ i}} \right) \right] \\ & = \Im \left[ \frac {e^{5^\circ i}-e^{180^\circ i}}{1-e^{5^\circ i}} \right] \\ & = \Im \left[ \frac {e^{5^\circ i}+1}{1-e^{5^\circ i}} \right] \\ & = \Im \left[ \frac {1+\cos 5^\circ + i \sin 5^\circ}{1-\cos 5^\circ - i \sin 5^\circ} \right] & \small \color{#3D99F6} \text{Multiply up and down with }1-\cos 5^\circ + i \sin 5^\circ \\ & = \Im \left[ \frac {2i \sin 5^\circ}{2-2\cos 5^\circ} \right] \\ & = \frac {\sin 5^\circ}{1-\cos 5^\circ} \\ & = \frac {2\sin 2.5^\circ \cos 2.5^\circ}{1-1 +2\sin^2 2.5^\circ} \\ & = \cot \frac {5^\circ}2 = \tan \frac {175^\circ}2 \end{aligned}

⟹ a + b = 175 + 2 = 177 \implies a+b = 175+2 = \boxed{177}

References:

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Nice solution, did the same, only different by one step, perhaps simpler. Instead of using the half-angle tangent substitution, maybe it is easier to put sin ⁡ ( 5 o ) = 2 sin ⁡ ( 2. 5 o ) cos ⁡ ( 2. 5 o ) \sin(5^o) = 2 \sin(2.5^o) \cos(2.5^o) and cos ⁡ ( 5 o ) = 1 − 2 sin ⁡ 2 ( 2. 5 o ) \cos(5^o) = 1 - 2 \sin^2(2.5^o)

Guilherme Niedu - 4 years ago

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Thanks, I will use this to shorten the solution.

Chew-Seong Cheong - 4 years ago

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Thank you for incorporating this in the solution.

Guilherme Niedu - 4 years ago

Also, when you multply below and above by the conjugate, the result is the imaginary part of 2 i sin ⁡ ( 5 o ) 2 − 2 cos ⁡ ( 5 o ) \frac{2i \sin(5^o)}{2 - 2 \cos(5^o)} . There is no 1 − cos ⁡ 2 ( 5 o ) 1 - \cos^2(5^o) above.

Guilherme Niedu - 4 years ago

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Thanks a lot.

Chew-Seong Cheong - 4 years ago

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You're welcome, sir.

Guilherme Niedu - 4 years ago

Done the same. Although i remember direct result of summation of sine and cosine when Angles r in AP . you derived it sir Upvoted!

Prakhar Bindal - 4 years ago
Ahmad Saad
May 16, 2017

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