Unfortunately, it's not in degrees

Geometry Level 5

∑ n = 1 ∞ sin ⁡ ( 360 n ) n = A B π − C \sum _{ n=1 }^{ \infty }{ \frac { \sin { \left( 360n \right) } }{ n } } =\frac { A }{ B } \pi -C

If the equation above holds true for positive integers A , B A,B and C C with A , B A,B coprime, submit your answer as C − B − A C-B-A .

Clarification : All the angles are measured in radians.


The answer is 63.

Jonas Katona by Jonas Katona , 22, USA

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1 solution

S = ∑ n = 1 ∞ sin ⁡ ( 360 n ) n = ∑ n = 1 ∞ ℑ ( e 360 n i ) n = ℑ ( ∑ n = 1 ∞ e 360 n i n ) = ℑ ( − ln ⁡ ( 1 − e 360 i ) ) = ℑ ( − ln ⁡ ( 1 − cos ⁡ 360 − i sin ⁡ 360 ) ) = ℑ ( − ln ⁡ ( ( 1 − cos ⁡ 360 ) 2 + sin ⁡ 2 360 exp ⁡ ( tan ⁡ − 1 ( − sin ⁡ 360 1 − cos ⁡ 360 ) i ) ) ) = ℑ ( − ln ⁡ ( ( 1 − cos ⁡ 360 ) 2 + sin ⁡ 2 360 ) + tan ⁡ − 1 ( sin ⁡ 360 1 − cos ⁡ 360 ) i ) = tan ⁡ − 1 ( sin ⁡ 360 1 − cos ⁡ 360 ) = tan ⁡ − 1 ( 2 tan ⁡ 180 1 + tan ⁡ 2 180 1 − 1 − tan ⁡ 2 180 1 + tan ⁡ 2 180 ) = tan ⁡ − 1 ( 2 tan ⁡ 180 1 + tan ⁡ 2 180 − 1 + tan ⁡ 2 180 ) = tan ⁡ − 1 ( 2 tan ⁡ 180 2 tan ⁡ 2 180 ) = tan ⁡ − 1 ( 1 tan ⁡ 180 ) = tan ⁡ − 1 ( cot ⁡ 180 ) = tan ⁡ − 1 ( tan ⁡ ( 115 π 2 − 180 ) ) = 115 π 2 − 180 \begin{aligned} S & = \sum_{n=1}^\infty \frac {\sin (360n)}n \\ & = \sum_{n=1}^\infty \frac {\Im \left(e^{360ni}\right)}n \\ & = \Im \left( \sum_{n=1}^\infty \frac {e^{360ni}}n \right) \\ & = \Im \left( -\ln \left(1-e^{360i} \right) \right) \\ & = \Im \left( -\ln \left(1-\cos 360 - i \sin 360 \right) \right) \\ & = \Im \left( -\ln \left(\sqrt{(1-\cos 360)^2+\sin^2 360}\exp \left( \tan^{-1 } \left( \frac {-\sin 360}{1-\cos 360} \right)i \right) \right) \right) \\ & = \Im \left( -\ln \left(\sqrt{(1-\cos 360)^2+\sin^2 360}\right) + \tan^{-1 } \left( \frac {\sin 360}{1-\cos 360} \right)i \right) \\ & = \tan^{-1 } \left( \frac {\sin 360}{1-\cos 360} \right) \\ & = \tan^{-1} \left( \frac {\frac {2 \tan 180}{1 + \tan^2 180}}{1 - \frac {1 - \tan^2 180}{1 + \tan^2 180}} \right) \\ & = \tan^{-1} \left( \frac {2\tan 180}{1 + \tan^2 180 - 1 + \tan^2 180} \right) \\ & = \tan^{-1} \left( \frac {2\tan 180}{2\tan^2 180} \right) \\ & = \tan^{-1} \left( \frac 1{\tan 180} \right) \\ & = \tan^{-1} \left( \cot 180 \right) \\ & = \tan^{-1} \left( \tan \left( \frac {115 \pi}2 - 180 \right) \right) \\ & = \frac {115 \pi}2 - 180 \end{aligned}

⟹ C − B − A = 180 − 2 − 115 = 63 \implies C-B-A = 180 - 2 - 115 = \boxed{63}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Hey, could you please tell me what is this capital J that you start using on the third equation?
Thank you very much.

A Former Brilliant Member - 4 years, 10 months ago

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That is actually an Im \text{Im} for imaginary part of a complex number . I actually key in " \ Im" and it turns out the be ℑ \Im in LaTex. The other is the Re ℜ \Re or real part of the complex number.

Chew-Seong Cheong - 4 years, 10 months ago

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