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Geometry Level 3

The average of the numbers n sin ⁡ ( n ∘ ) n \sin (n^{\circ}) for n = 2 , 4 , 6 , … , 180 n=2,4,6,\ldots , 180 is?

1 tan ⁡ 1 ∘ \tan 1^{\circ} cot ⁡ 1 ∘ \cot 1^{\circ} 0

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Prashant Kr by Prashant Kr , 22, India

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2 solutions

Mas Mus
Sep 11, 2015

Let X ‾ \overline{X} is the average of the numbers.

X ‾ = 2 sin ⁡ 2 ∘ + 4 sin ⁡ 4 ∘ + 6 sin ⁡ 6 ∘ + … + 174 sin ⁡ 17 4 ∘ + 176 sin ⁡ 17 6 ∘ + 178 sin ⁡ 17 8 ∘ + 180 sin ⁡ 18 0 ∘ 90 \overline{X}=\large{\frac{2\sin2^{\circ}+4\sin4^{\circ}+6\sin6^{\circ}+~\dots~+174\sin174^{\circ}+176\sin176^{\circ}+178\sin178^{\circ}+180\sin180^{\circ}}{90}}

Using the identity, sin ⁡ ( 180 − α ) = sin ⁡ α ~\sin(180-\alpha)=\sin\alpha , we get that

sin ⁡ 2 ∘ = sin ⁡ 17 8 ∘ , sin ⁡ 4 ∘ = sin ⁡ 17 6 ∘ , sin ⁡ 4 ∘ = sin ⁡ 17 4 ∘ , \sin2^{\circ}=\sin178^{\circ},~~\sin4^{\circ}=\sin176^{\circ},~~\sin4^{\circ}=\sin174^{\circ}, and soon, so we can simplfy the numerator become:

X ‾ = 180 ( sin ⁡ 2 ∘ + sin ⁡ 4 ∘ + sin ⁡ 6 ∘ + … + sin ⁡ 8 4 ∘ + sin ⁡ 8 6 ∘ + sin ⁡ 8 8 ∘ ) + 90 sin ⁡ 9 0 ∘ + 180 sin ⁡ 18 0 ∘ 90 = 2 cos ⁡ 8 8 ∘ ( sin ⁡ 2 ∘ + sin ⁡ 4 ∘ + sin ⁡ 6 ∘ + … + sin ⁡ 8 4 ∘ + sin ⁡ 8 6 ∘ + sin ⁡ 8 8 ∘ ) cos ⁡ 8 8 ∘ + 1 \begin{aligned}\overline{X}&=\frac{180\left(\sin2^{\circ}+\sin4^{\circ}+\sin6^{\circ}+~\dots~+\sin84^{\circ}+\sin86^{\circ}+\sin88^{\circ}\right)+90\sin90^{\circ}+180\sin180^{\circ}}{90}\\&=\frac{2\cos88^{\circ}\left(\sin2^{\circ}+\sin4^{\circ}+\sin6^{\circ}+~\dots~+\sin84^{\circ}+\sin86^{\circ}+\sin88^{\circ}\right)}{\cos88^{\circ}}+1\end{aligned}

Now, multiply all of items on bracket by cos ⁡ 8 8 ∘ \cos88^{\circ} and use 2 cos ⁡ α sin ⁡ β = sin ⁡ ( α + β ) − sin ⁡ ( α − β ) ~2\cos\alpha~\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta)

2 cos ⁡ 8 8 ∘ sin ⁡ 2 ∘ = sin ⁡ 9 0 ∘ − sin ⁡ 8 6 ∘ = sin ⁡ 9 0 ∘ − sin ⁡ 9 4 ∘ 2\cos88^{\circ}~\sin2^{\circ}=\sin90^{\circ}-\sin86^{\circ}=\sin90^{\circ}-\color{#D61F06}{\sin94^{\circ}}

2 cos ⁡ 8 8 ∘ sin ⁡ 4 ∘ = sin ⁡ 9 2 ∘ − sin ⁡ 8 4 ∘ = sin ⁡ 9 2 ∘ − sin ⁡ 9 6 ∘ 2\cos88^{\circ}~\sin4^{\circ}=\sin92^{\circ}-\sin84^{\circ}=\sin92^{\circ}-\color{#3D99F6}{\sin96^{\circ}}

2 cos ⁡ 8 8 ∘ sin ⁡ 6 ∘ = sin ⁡ 9 4 ∘ − sin ⁡ 8 2 ∘ = sin ⁡ 9 4 ∘ − sin ⁡ 9 8 ∘ 2\cos88^{\circ}~\sin6^{\circ}=\sin94^{\circ}-\sin82^{\circ}=\color{#D61F06}{\sin94^{\circ}}-\color{#20A900}{\sin98^{\circ}}

2 cos ⁡ 8 8 ∘ sin ⁡ 8 ∘ = sin ⁡ 9 6 ∘ − sin ⁡ 8 0 ∘ = sin ⁡ 9 6 ∘ − sin ⁡ 10 0 ∘ 2\cos88^{\circ}~\sin8^{\circ}=\sin96^{\circ}-\sin80^{\circ}=\color{#3D99F6}{\sin96^{\circ}}-\color{#BA33D6}{\sin100^{\circ}}

⋮ ~~\vdots

2 cos ⁡ 8 8 ∘ sin ⁡ 8 0 ∘ = sin ⁡ 16 8 ∘ − sin ⁡ 8 ∘ = sin ⁡ 16 8 ∘ − sin ⁡ 17 2 ∘ 2\cos88^{\circ}~\sin80^{\circ}=\sin168^{\circ}-\sin8^{\circ}=\color{grey}{\sin168^{\circ}}-\color{#69047E}{\sin172^{\circ}}

2 cos ⁡ 8 8 ∘ sin ⁡ 8 2 ∘ = sin ⁡ 17 0 ∘ − sin ⁡ 6 ∘ = sin ⁡ 17 0 ∘ − sin ⁡ 17 4 ∘ 2\cos88^{\circ}~\sin82^{\circ}=\sin170^{\circ}-\sin6^{\circ}=\color{silver}{\sin170^{\circ}}-\color{#EC7300}{\sin174^{\circ}}

2 cos ⁡ 8 8 ∘ sin ⁡ 8 4 ∘ = sin ⁡ 17 2 ∘ − sin ⁡ 4 ∘ = sin ⁡ 17 2 ∘ − sin ⁡ 17 6 ∘ 2\cos88^{\circ}~\sin84^{\circ}=\sin172^{\circ}-\sin4^{\circ}=\color{#69047E}{\sin172^{\circ}}-\color{#E81990}{\sin176^{\circ}}

2 cos ⁡ 8 8 ∘ sin ⁡ 8 6 ∘ = sin ⁡ 17 4 ∘ − sin ⁡ 2 ∘ = sin ⁡ 17 4 ∘ − sin ⁡ 2 ∘ 2\cos88^{\circ}~\sin86^{\circ}=\sin174^{\circ}-\sin2^{\circ}=\color{#EC7300}{\sin174^{\circ}}-\sin2^{\circ}

2 cos ⁡ 8 8 ∘ sin ⁡ 8 8 ∘ = sin ⁡ 17 6 ∘ − sin ⁡ 0 ∘ = sin ⁡ 17 6 ∘ − sin ⁡ 0 ∘ 2\cos88^{\circ}~\sin88^{\circ}=\sin176^{\circ}-\sin0^{\circ}=\color{#E81990}{\sin176^{\circ}}-\sin0^{\circ}

All items that have the same color will cancel each other and the black items are remaining. Therefore, we can re-write the last equation of X ‾ \overline{X} as

X ‾ = sin ⁡ 9 0 ∘ + sin ⁡ 9 2 ∘ − sin ⁡ 2 ∘ cos ⁡ 8 8 ∘ + 1 = 2 sin ⁡ 9 1 ∘ cos ⁡ 1 ∘ − cos ⁡ 8 8 ∘ cos ⁡ 8 8 ∘ + 1 = 2 sin ⁡ 9 1 ∘ cos ⁡ 1 ∘ 2 sin ⁡ 1 ∘ cos ⁡ 1 ∘ = 2 cos ⁡ 1 ∘ cos ⁡ 1 ∘ 2 sin ⁡ 1 ∘ cos ⁡ 1 ∘ = cot ⁡ 1 ∘ \begin{aligned}\overline{X}&=\frac{\sin90^{\circ}+\sin92^{\circ}-\sin2^{\circ}}{\cos88^{\circ}}+1=\frac{2\sin91^{\circ}\cos1^{\circ}-\cos88^{\circ}}{\cos88^{\circ}}+1\\\\&=\frac{2\sin91^{\circ}\cos1^{\circ}}{2\sin1^{\circ} \cos1^{\circ}}=\frac{2\cos1^{\circ}\cos1^{\circ}}{2\sin1^{\circ} \cos1^{\circ}}=\cot1^{\circ}\end{aligned}

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Although this is not a proper way of solving it, it is a multiple choice question so we can use the process of elimination. sin(180-x) = sin(x), so for 0<x<180, sinx>0 and sin180=0. So all of the numbers are non-negative. 2+4+6+8+...+180 = 2(0.5)(90+91)= 8190. There are 90 terms so the average of the coefficients is 90. The middle positive value of sin x is 0.5, but more than half of the values for x will have a sine greater than 0.5 (30<x<150). So the average will be greater than 90(0.5), which is 45. Thus, the average of the numbers will be above 45. Check each option: 0 is under 45, 1 is under 45, tan 1 is less than 1, but cot 1 is greater that 45. (Correct to 2 decimal places, cot1= 57.29). So the average of the numbers is cot1.

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Moderator note:

This solution has been marked wrong as it's basically an antisolution.

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