Summation + Trigonometry

Geometry Level 3

∑ n = 1 18 cos ⁡ 2 ( 5 n ∘ ) \large\sum _{n=1}^{18} \cos ^{ 2 }(5n^{\circ})

Find the value of the summation above.

If the answer is of the form a b \frac{a}{b} , where a a and b b are positive coprime integers, find a + b a+b .


The answer is 19.

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Vaibhav Prasad by Vaibhav Prasad , 21, India

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4 solutions

Adarsh Kumar
Jun 22, 2015

We have that cos ⁡ 2 5 = c o s 2 5 cos ⁡ 2 85 = sin ⁡ 2 5 cos ⁡ 2 10 = cos ⁡ 2 10 cos ⁡ 2 80 = sin ⁡ 2 10 . . . cos ⁡ 2 40 = cos ⁡ 2 40 cos ⁡ 2 50 = sin ⁡ 2 40 \cos^2 5=cos^2 5\\ \cos^2 85=\sin^2 5 \\ \cos^2 10=\cos^2 10\\ \cos^2 80=\sin^2 10\\ .\\ .\\ .\\ \cos^2 40=\cos^2 40\\ \cos^2 50=\sin^2 40 .Sum of these would be 1 + 1 + 1 + . . . . + 1 = 8 1+1+1+....+1=8 and cos ⁡ 2 45 = 1 2 \cos^2 45=\dfrac{1}{2} ,hence sum would be 8 + 1 2 = 17 2 8+\dfrac{1}{2}=\dfrac{17}{2} .Hence sum is 19 19 .And done!

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Moderator note:

Good pairing up approach.

As always, if the answer is "very nice", think about "How could one get to this very nice answer"?

I overthought it, it seems. Great solution!

Jake Lai - 5 years, 11 months ago
Jake Lai
Jun 22, 2015

Let our sum be S S . The double-angle identity for cosine is 2 cos ⁡ 2 θ − 1 ≡ cos ⁡ ( 2 θ ) 2\cos^{2} \theta -1 \equiv \cos(2\theta) . Using this,

2 S − 18 = ∑ n = 1 18 ( 2 cos ⁡ 2 ( 5 n ∘ ) − 1 ) = ∑ n = 1 18 cos ⁡ ( n π / 18 ) 2S-18 = \sum_{n=1}^{18} \left( 2\cos^{2}(5n^{\circ})-1 \right) = \sum_{n=1}^{18} \cos(n\pi/18)

It is well-known that ∑ n = 1 N cos ⁡ ( n π / N ) = − 1 \displaystyle \sum_{n=1}^{N} \cos(n\pi/N) = -1 ; this is left as an exercise for the reader.

Thus, we have 2 S − 18 = − 1 2S-18 = -1 and so S = 17 2 S = \boxed{\frac{17}{2}} .

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Moderator note:

Bonus question : Prove that ∑ n = 1 18 cos ⁡ 4 ( 5 n ∘ ) = 25 4 \displaystyle \sum_{n=1}^{18} \cos^4(5n^\circ) = \frac{25}4 .

[Reply to Challenge Master Note]

We make excessive use of the complementary angle formula sin ⁡ ( x ) = cos ⁡ ( 9 0 ∘ − x ) \sin(x)=\cos(90^\circ -x) and the identity sin ⁡ 2 ( x ) + cos ⁡ 2 ( x ) = 1 \sin^2(x)+\cos^2(x)=1 and also use the double angle formula 2 sin ⁡ ( x ) cos ⁡ ( x ) = sin ⁡ ( 2 x ) 2\sin(x)\cos(x)=\sin(2x) . Keeping track of them is left as an exercise to the reader.

∑ n = 1 18 cos ⁡ 4 ( 5 n ∘ ) = cos ⁡ 4 ( 4 5 ∘ ) + cos ⁡ 4 ( 9 0 ∘ ) + ∑ n = 1 8 ( cos ⁡ 4 ( 5 n ∘ ) + sin ⁡ 4 ( 5 n ∘ ) ) = 1 4 + 0 + ∑ n = 1 8 ( ( cos ⁡ 2 ( 5 n ∘ ) + sin ⁡ 2 ( 5 n ∘ ) ) 2 − 1 2 sin ⁡ 2 ( 10 n ∘ ) ) = 1 4 + ∑ n = 1 8 ( 1 − 1 2 sin ⁡ 2 ( 10 n ∘ ) ) = 1 4 + 8 − 1 2 ∑ n = 1 4 ( sin ⁡ 2 ( 10 n ∘ ) + cos ⁡ 2 ( 10 n ∘ ) ) = 1 4 + 8 − 1 2 ∑ n = 1 4 1 = 1 4 + 8 − 1 2 × 4 = 25 4 \begin{aligned}\sum_{n=1}^{18}\cos^4(5n^\circ)&=\cos^4(45^\circ)+\cos^4(90^\circ)+\sum_{n=1}^{8}\left(\cos^4(5n^\circ)+\sin^4(5n^\circ)\right)\\&=\frac{1}{4}+0+\sum_{n=1}^8\left(\left(\cos^2(5n^\circ)+\sin^2(5n^\circ)\right)^2-\frac 12\sin^2(10n^\circ)\right)\\&=\frac 14+\sum_{n=1}^8\left(1-\frac 12\sin^2(10n^\circ)\right)\\&=\frac 14+8-\frac 12\sum_{n=1}^4\left(\sin^2(10n^\circ)+\cos^2(10n^\circ)\right)\\&=\frac 14+8-\frac 12\sum_{n=1}^4 1=\frac 14+8-\frac 12\times 4=\frac{25}{4}\end{aligned}

There might be another proof using a method similar to what Jake did in his original solution but I think this is the simplest proof possible.

Prasun Biswas - 5 years, 11 months ago
Chew-Seong Cheong
Jun 22, 2015

∑ n = 1 18 cos ⁡ 2 5 n ∘ = 1 2 ∑ n = 1 18 ( 2 cos ⁡ 2 5 n ∘ − 1 + 1 ) = 1 2 ∑ n = 1 18 ( cos ⁡ 10 n ∘ + 1 ) = 1 2 ∑ n = 1 8 ( cos ⁡ 10 n ∘ + cos ⁡ ( 180 − 10 n ) ∘ ) + 1 2 cos ⁡ 18 0 ∘ + 1 2 ( 18 ) = 1 2 ∑ n = 1 8 ( cos ⁡ 10 n ∘ − cos ⁡ 10 n ∘ ) + 1 2 ( − 1 ) + 9 = 0 − 1 2 + 9 = 17 2 ⇒ a + b = 17 + 2 = 19 \begin{aligned} \sum_{n=1}^{18} \cos^2{5n^\circ} & = \frac{1}{2} \sum_{n=1}^{18} \left(2 \cos^2{5n^\circ} - 1 + 1\right) \\ & = \frac{1}{2} \sum_{n=1}^{18} \left(\cos{10n^\circ} + 1\right) \\ & = \frac{1}{2} \sum_{n=1}^{8} \left(\cos{10n^\circ} + \cos{(180-10n)^\circ} \right) + \frac{1}{2} \cos{180^\circ} + \frac{1}{2} (18) \\ & = \frac{1}{2} \sum_{n=1}^{8} \left(\cos{10n^\circ} - \cos { 10 n^\circ} \right) + \frac{1}{2}(-1) + 9\\ & = 0 - \frac{1}{2} + 9 \\ & = \frac{17}{2} \\ \Rightarrow a + b & = 17 + 2 = \boxed{19} \end{aligned}

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Moderator note:

Nice use of the supplementary angles formula: cos ⁡ ( A ) = − cos ⁡ ( 18 0 ∘ − A ) \cos(A) = -\cos(180^\circ - A) .

Rwit Panda
Jun 23, 2015

The series becomes cos^{2}(5) + cos^{2}(10) + cos^{2}(15)...........+cos^{2}(90). cos(85)=sin(5). Thus cos^{2}(5)+cos^{2}(85)=1. There exist 8 such sets. cos^{2}(45)=0.5 and cos^{2}(90)=0 Thus the summation equals 8.5

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I did same!!

Dev Sharma - 5 years, 7 months ago

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