Trigonometric Identity Maybe?

Geometry Level 4

I made an elementary trigonometric identity complicated to get this expression.

∑ k = 1 10 ( sin ⁡ ( 2 π k 11 ) − i cos ⁡ ( 2 π k 11 ) ) \large \sum_{k=1}^{10} \left ( \sin \left ( \frac {2\pi k}{11 } \right ) - i \cos \left ( \frac {2\pi k}{11 } \right ) \right )

What is the value of 4 times the square of the expression above?

Details and Assumptions : i = − 1 i =\sqrt {-1} denotes the imaginary unit .

Also try this .


The answer is -4.00.

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Tanishq Varshney by Tanishq Varshney , 23, India

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3 solutions

Chew-Seong Cheong
Feb 10, 2018

By Euler's formula we have e i x = cos ⁡ x + i sin ⁡ x e^{ix} = \cos x + i \sin x . Then

S = ∑ k = 1 10 sin ⁡ ( 2 π k 11 ) − i cos ⁡ ( 2 π k 11 ) = ∑ k = 1 10 − i e 2 π k 11 = ∑ k = 1 10 − i ω k where ω = e 2 π 11 is the 11th root of unit. = − i ( ∑ k = 0 10 ω k − 1 ) Note that ∑ k = 0 10 ω k = 0 = i \begin{aligned} S & = \sum_{k=1}^{10} \sin \left(\dfrac {2\pi k}{11} \right) - i \cos \left(\dfrac {2\pi k}{11} \right) \\ & = \sum_{k=1}^{10} - i {\color{#3D99F6} e^{\frac {2\pi k}{11}}} = \sum_{\color{#3D99F6}k=1}^{10} - i \color{#3D99F6} \omega^k & \small \color{#3D99F6} \text{where } \omega = e^{\frac {2\pi}{11}} \text{ is the 11th root of unit.} \\ & = -i \left(\sum_{\color{#D61F06}k=0}^{10} \omega^k - 1 \right) & \small \color{#3D99F6} \text{Note that }\sum_{\color{#D61F06}k=0}^{10} \omega^k = 0 \\ & = i \end{aligned}

Therefore 4 S 2 = 4 ( i 2 ) = − 4 4S^2 = 4(i^2) = \boxed{-4} .

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Small typo: Note thar \text{Note thar}

Bernard Peh - 3 years, 4 months ago

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Thanks, corrected.

Chew-Seong Cheong - 3 years, 4 months ago

@Chew-Seong Cheong , nice solution sir. Also, I am missing your solutions to my problems. I always look forward to see your solutions for my problems.

Hana Wehbi - 3 years, 4 months ago
Nishant Rai
Apr 5, 2015

simpler one. Take -i common from the whole expression and then evaluate sigma from k=0. To compensate for the changes, subtract -1. Once you start the sigma part from k=0 , one can easily see that it is sum of the roots of unity, which is equal to 0. so you are left with only -1 in the bracket. And the expression evaluates to 'i'.

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take − i -i common, the expression becomes

− i ( ∑ k = 1 10 c o s ( 2 k π 11 ) + i s i n ( 2 k π 11 ) ) -i (\displaystyle \sum_{k=1}^{10} cos(\frac{2k \pi}{11})+i sin(\frac{2k \pi}{11}))

− i ∑ k = 1 10 e i 2 k π 11 -i \displaystyle \sum_{k=1}^{10} e^{i \frac{2 k \pi}{11}} ..... ( 1 ) (1)

11 roots of unity 1 , x 1 , x 2 , x 3 . . . . . . x 10 1,x_{1},x_{2},x_{3}......x_{10}

Also 1 + x 1 + x 2 + . . . . x 10 = 0 1+x_{1}+x_{2}+....x_{10}=0

x 1 + x 2 + . . . . x 10 = − 1 x_{1}+x_{2}+....x_{10}=-1

in ( 1 ) (1)

− i ( x 1 + x 2 + x 3 + . . . . . x 10 ) -i (x_{1}+x_{2}+x_{3}+.....x_{10})

i \boxed{i}

Tanishq Varshney - 6 years, 2 months ago
Jerry Bao
Feb 14, 2018

Think of each part of the sum as a single vector. These vectors form the vertices of a regular hendecagon. But one of the vectors are missing from the sum. This is the case when k=11. Plug k=11 to get -i. Due to the fact that it is a regular hendecagon, the other vectors must cancel out the k=11 vector, so their sum must equal to i.

Squaring and multiplying by 4 gives -4.

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