A Trigonometric sequence (No calculators!)

Geometry Level 3

Evaluate

sin ⁡ 2 ( 1 ∘ ) + sin ⁡ 2 ( 2 ∘ ) + sin ⁡ 2 ( 3 ∘ ) + . . . + sin ⁡ 2 ( 8 9 ∘ ) + sin ⁡ 2 ( 9 0 ∘ ) \sin^2(1^\circ)+\sin^2(2^\circ)+\sin^2(3^\circ)+...+\sin^2(89^\circ)+\sin^2(90^\circ)


The answer is 45.5.

Tarmo Taipale by Tarmo Taipale , 22, Finland

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2 solutions

Chew-Seong Cheong
Jul 27, 2017

S = sin ⁡ 2 1 ∘ + sin ⁡ 2 2 ∘ + sin ⁡ 2 3 ∘ + ⋯ + sin ⁡ 2 8 9 ∘ + sin ⁡ 2 9 0 ∘ = ∑ n = 1 90 sin ⁡ 2 n ∘ = 1 2 ( ∑ n = 1 90 sin ⁡ 2 n ∘ + ∑ n = 1 90 sin ⁡ 2 n ∘ ) = 1 2 ( ∑ n = 1 90 sin ⁡ 2 n ∘ + ∑ n = 1 90 cos ⁡ 2 ( 90 − n ) ∘ ) = 1 2 ( ∑ n = 1 90 sin ⁡ 2 n ∘ + ∑ n = 0 89 cos ⁡ 2 n ∘ ) = 1 2 ( ∑ n = 1 89 ( sin ⁡ 2 n ∘ + cos ⁡ 2 n ∘ ) + sin ⁡ 9 0 ∘ + cos ⁡ 0 ∘ ) = 1 2 ( 89 + 1 + 1 ) = 45.5 \begin{aligned} S & = \sin^2 1^\circ + \sin^2 2^\circ + \sin^2 3^\circ + \cdots + \sin^2 89^\circ + \sin^2 90^\circ \\ & = \sum_{n=1}^{90} \sin^2 n^\circ \\ & = \frac 12 \left(\sum_{n=1}^{90} \sin^2 n^\circ + \sum_{n=1}^{90} \sin^2 n^\circ \right) \\ & = \frac 12 \left(\sum_{\color{#3D99F6}n=1}^{\color{#3D99F6}90} \sin^2 n^\circ + \sum_{\color{#3D99F6}n=1}^{\color{#3D99F6}90} \cos^2 (90- n)^\circ \right) \\ & = \frac 12 \left(\sum_{\color{#3D99F6}n=1}^{\color{#3D99F6}90} \sin^2 n^\circ + \sum_{\color{#D61F06}n=0}^{\color{#D61F06}89} \cos^2 n^\circ \right) \\ & = \frac 12 \left(\sum_{\color{#3D99F6}n=1}^{\color{#D61F06}89} \left(\sin^2 n^\circ + \cos^2 n^\circ\right) + {\color{#3D99F6} \sin 90^\circ} + {\color{#D61F06}\cos 0^\circ} \right) \\ & = \frac 12 \left(89 + {\color{#3D99F6} 1} + {\color{#D61F06}1} \right) = \boxed{45.5} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Tarmo Taipale
Jul 26, 2017

By sin ⁡ ( 9 0 ∘ − x ) = cos ⁡ x \sin(90^\circ-x)=\cos x we get

sin ⁡ 2 ( 1 ∘ ) + sin ⁡ 2 ( 2 ∘ ) + sin ⁡ 2 ( 3 ∘ ) . . . + sin ⁡ 2 ( 8 9 ∘ ) + sin ⁡ 2 ( 9 0 ∘ ) = sin ⁡ 2 ( 1 ∘ ) + cos ⁡ 2 ( 1 ∘ ) + sin ⁡ 2 ( 2 ∘ ) + cos ⁡ 2 ( 2 ∘ ) + . . . + sin ⁡ 2 ( 4 4 ∘ ) + cos ⁡ 2 ( 4 4 ∘ ) + sin ⁡ 2 ( 9 0 ∘ ) + sin ⁡ 2 ( 4 5 ∘ ) \sin^2(1^\circ)+\sin^2(2^\circ)+\sin^2(3^\circ)...+\sin^2(89^\circ)+\sin^2(90^\circ) = \sin^2(1^\circ)+\cos^2(1^\circ)+\sin^2(2^\circ)+\cos^2(2^\circ)+...+\sin^2(44^\circ)+\cos^2(44^\circ)+\sin^2(90^\circ)+\sin^2(45^\circ)

By sin ⁡ 2 x + cos ⁡ 2 x = 1 \sin^2x+\cos^2x=1 , sin ⁡ 9 0 ∘ = 1 \sin 90^\circ = 1 and sin ⁡ 4 5 ∘ = 1 2 \sin 45^\circ = \frac{1}{\sqrt{2}} we get

sin ⁡ 2 ( 1 ∘ ) + cos ⁡ 2 ( 1 ∘ ) + sin ⁡ 2 ( 2 ∘ ) + cos ⁡ 2 ( 2 ∘ ) + . . . + sin ⁡ 2 ( 4 4 ∘ ) + cos ⁡ 2 ( 4 4 ∘ ) + sin ⁡ 2 ( 9 0 ∘ ) + sin ⁡ 2 ( 4 5 ∘ ) = 44 + 1 + 1 2 \sin^2(1^\circ)+\cos^2(1^\circ)+\sin^2(2^\circ)+\cos^2(2^\circ)+...+\sin^2(44^\circ)+\cos^2(44^\circ)+\sin^2(90^\circ)+\sin^2(45^\circ) = 44+1+\frac{1}{2}

44 + 1 + 1 2 = 45.5 44+1+\frac{1}{2}=\boxed{45.5}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Your solution seems perfect but for latex coding i will tell you to use \sin(x) as sine is a function. It will give an output sin ⁡ ( x ) \sin(x) where your output was s i n ( x ) sin(x) . I think you got the difference.

Md Zuhair - 3 years, 10 months ago

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Yeah I understand, maybe it would look better.

Tarmo Taipale - 3 years, 10 months ago

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Now it's fixed.

Tarmo Taipale - 3 years, 10 months ago

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