Sine Party!

Geometry Level 3

sin 1 sin 4 5 sin 4 6 + sin 1 sin 4 7 sin 4 8 + sin 1 sin 4 9 sin 5 0 + + sin 1 sin 13 1 sin 13 2 + sin 1 sin 13 3 sin 13 4 = ? \displaystyle\frac{\sin 1^\circ}{\sin 45^\circ \sin 46^\circ} + \frac{\sin 1^\circ}{\sin 47^\circ \sin 48^\circ} + \frac{\sin 1^\circ}{\sin 49^\circ \sin 50^\circ} + \cdots \\\displaystyle \cdots + \frac{\sin 1^\circ}{\sin 131^\circ \sin 132^\circ} + \frac{\sin 1^\circ}{\sin 133^\circ \sin 134^\circ} =\ ?


Source : 103 Trigonometry Problems, by Titu Andreescu.


The answer is 1.

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1 solution

Chew-Seong Cheong
Apr 27, 2016

S = sin 1 sin 4 5 sin 4 6 + sin 1 sin 4 7 sin 4 8 + sin 1 sin 4 9 sin 5 0 + . . . + sin 1 sin 13 3 sin 13 4 = n = 0 44 sin 1 sin ( 45 + 2 n ) sin ( 45 + 2 n + 1 ) = n = 0 44 sin ( 45 + 2 n + 1 45 + 2 n ) sin ( 45 + 2 n ) sin ( 45 + 2 n + 1 ) = n = 0 44 sin ( 45 + 2 n + 1 ) cos ( 45 + 2 n ) sin ( 45 + 2 n ) cos ( 45 + 2 n + 1 ) sin ( 45 + 2 n ) sin ( 45 + 2 n + 1 ) = n = 0 44 ( cot ( 45 + 2 n ) cot ( 45 + 2 n + 1 ) ) = n = 45 134 ( 1 ) n + 1 cot n = n = 45 89 ( 1 ) n + 1 cot n cot 9 0 + n = 91 134 ( 1 ) n + 1 cot n As cot ( 18 0 θ ) = cot θ = n = 45 89 ( 1 ) n + 1 cot n 0 + n = 91 134 ( 1 ) n cot ( 180 n ) = n = 45 89 ( 1 ) n + 1 cot n + n = 46 89 ( 1 ) n cot n = cot 4 5 = 1 \begin{aligned} S & = \frac{\sin 1^\circ}{\sin 45^\circ \sin 46^\circ} + \frac{\sin 1^\circ}{\sin 47^\circ \sin 48^\circ} + \frac{\sin 1^\circ}{\sin 49^\circ \sin 50^\circ} + ... + \frac{\sin 1^\circ}{\sin 133^\circ \sin 134^\circ} \\ & = \sum_{n=0}^{44} \frac{\sin 1^\circ}{\sin (45+2n)^\circ \sin (45 +2n+1)^\circ} \\ & = \sum_{n=0}^{44} \frac{\sin (45+2n+1 - 45+2n)^\circ}{\sin (45+2n)^\circ \sin (45 +2n+1)^\circ} \\ & = \sum_{n=0}^{44} \frac{\sin (45+2n+1)^\circ \cos (45+2n)^\circ - \sin (45+2n)^\circ \cos (45+2n+1)^\circ}{\sin (45+2n)^\circ \sin (45 +2n+1)^\circ} \\ & = \sum_{n=0}^{44} (\cot (45+2n)^\circ - \cot (45+2n+1)^\circ) \\ & = \sum_{n=45}^{134} (-1)^{n+1}\cot n^\circ \\ & = \sum_{n=45}^{89} (-1)^{n+1}\cot n^\circ - \cot 90^\circ + \color{#3D99F6}{\sum_{n=91}^{134} (-1)^{n+1}\cot n^\circ} \quad \quad \small \color{#3D99F6}{\text{As }\cot (180^\circ-\theta) = - \cot \theta} \\ & = \sum_{n=45}^{89} (-1)^{n+1}\cot n^\circ - 0 + \color{#3D99F6}{\sum_{n=91}^{134} (-1)^{n}\cot (180-n)^\circ} \\ & = \sum_{n=45}^{89} (-1)^{n+1}\cot n^\circ + \color{#3D99F6}{\sum_{n=46}^{89} (-1)^{n}\cot n^\circ} \\ & = \cot 45^\circ \\ & = \boxed{1} \end{aligned}

Exactly the same solution!!

Aakash Khandelwal - 5 years, 1 month ago

This is from "103 Trigonometry Problems" by Titu Andreescu.

Shourya Pandey - 5 years, 1 month ago

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