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Geometry Level 4

∑ k = 1 15 cis 2 k − 1 ( π 36 ) \large\sum_{k=1}^{15} \textrm{cis}^{2k-1}\left( \frac{\pi}{36} \right)

Compute the imaginary part of the summation above.

Notation

  • cis θ = cos ⁡ θ + i sin ⁡ θ \textrm{cis }\theta = \cos \theta + i \sin\theta .
1 4 sin ⁡ π 36 \frac{1}{4 \sin \frac{\pi}{36}} sin ⁡ π 36 \sin \frac{\pi}{36} 1 4 \frac{1}{4} 2 + 3 4 sin ⁡ π 36 \frac{2+\sqrt{3}}{4 \sin \frac{\pi}{36}} 2 − 3 4 sin ⁡ π 36 \frac{2-\sqrt{3}}{4 \sin \frac{\pi}{36}}

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Giancarlo Miragliotta by Giancarlo Miragliotta , 56, Brazil

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3 solutions

First note that

cis θ = cos ⁡ θ + i sin ⁡ θ ⇒ ⇒ cis 2 k − 1 θ = cos ⁡ [ ( 2 k − 1 ) θ ] + i sin ⁡ [ ( 2 k − 1 ) θ ] . \begin{aligned} \textrm{cis } \theta & = \cos \theta + i \sin\theta \Rightarrow \\ \Rightarrow \textrm{cis }^{2k-1} \theta & = \cos [( 2k-1) \theta] + i \sin [( 2k-1 ) \theta]. \\ \end{aligned}

And then

S = ∑ k = 1 15 cis 2 k − 1 ( π 36 ) = ∑ k = 1 15 cis [ ( 2 k − 1 ) π 36 ] . \begin{aligned} S & = \large\sum_{k=1}^{15} \textrm{cis}^{2k-1}\left( \frac{\pi}{36} \right) \\ & = \large\sum_{k=1}^{15} \textrm{cis} \left[ \frac{(2k-1)\pi}{36} \right]. \\ \end{aligned}

If Im [ S ] \textrm{Im} [S] is the imaginary part of the S S then

Im [ S ] = ∑ k = 1 15 sin ⁡ [ ( 2 k − 1 ) π 36 ] . = sin ⁡ π 36 + sin ⁡ 3 π 36 + … + sin ⁡ 29 π 36 . \begin{aligned} \textrm{Im} [S] & = \large \sum_{k=1}^{15} \sin \left[ \frac{(2k-1)\pi}{36} \right]. \\ & = \sin \frac{\pi}{36} + \sin \frac{3\pi}{36} + \ldots +\sin \frac{29\pi}{36}.\\ \end{aligned}

Multiplying both the sides by − 2 sin ⁡ π 36 -2\sin \frac{\pi}{36}

− 2 sin ⁡ π 36 Im [ S ] = − 2 sin ⁡ π 36 sin ⁡ π 36 − 2 sin ⁡ π 36 sin ⁡ 3 π 36 − … − 2 sin ⁡ π 36 sin ⁡ 29 π 36 = − 2 sin ⁡ 2 π 36 + [ cos ⁡ 2 π 18 − cos ⁡ π 18 ] + [ cos ⁡ 3 π 18 − cos ⁡ 2 π 18 ] + … … + [ cos ⁡ 15 π 18 − cos ⁡ 14 π 18 ] = − 2 sin ⁡ 2 π 36 − cos ⁡ π 18 + cos ⁡ 15 π 18 = − 2 sin ⁡ 2 π 36 − cos ⁡ 2 π 36 + cos ⁡ 5 π 6 = − 2 sin ⁡ 2 π 36 − ( 1 − 2 sin ⁡ 2 π 36 ) + cos ⁡ 5 π 6 . \begin{aligned} -2\sin \frac{\pi}{36} \textrm{Im} [S] & = -2\sin \frac{\pi}{36} \sin \frac{\pi}{36} -2\sin \frac{\pi}{36} \sin \frac{3\pi}{36} - \ldots -2\sin \frac{\pi}{36} \sin \frac{29\pi}{36}\\ & =-2\sin^2 \frac{\pi}{36} + \biggl[ \cos \frac{2\pi}{18}-\cos \frac{\pi}{18} \biggl] +\biggl[ \cos \frac{3\pi}{18}-\cos \frac{2\pi}{18} \biggl] +\ldots\\ & \ldots +\biggl[ \cos \frac{15\pi}{18}-\cos \frac{14\pi}{18} \biggl] \\ & =-2\sin^2 \frac{\pi}{36} - \cos \frac{\pi}{18} + \cos \frac{15\pi}{18}\\ & =-2\sin^2 \frac{\pi}{36} - \cos \frac{2\pi}{36} + \cos \frac{5\pi}{6}\\ & =-2\sin^2 \frac{\pi}{36} - (1-2\sin^2 \frac{\pi}{36}) + \cos \frac{5\pi}{6}.\\ \end{aligned}

Thus

− 2 sin ⁡ π 36 Im [ S ] = − 1 − 3 2 Im [ S ] = 2 + 3 4 sin ⁡ π 36 . \begin{aligned} -2\sin \frac{\pi}{36} \textrm{Im} [S] & =-1 - \frac{\sqrt3}{2}\\ \textrm{Im} [S] & = \boxed{\dfrac{2+\sqrt{3}}{4 \sin \frac{\pi}{36}}}.\\ \end{aligned}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

How did you get the motivation to multiply both sides by -2sin?

Roni Jaira - 5 years, 5 months ago

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Roni... try this problem:

https://brilliant.org/problems/day-7-seven-terms-in-a-tricky-trig-sum/?group=R7BPCpcARbs3&ref_id=1049242

;-)

Giancarlo Miragliotta - 5 years, 5 months ago
Chew-Seong Cheong
Dec 24, 2015

We note that cis θ = c o s θ + i sin ⁡ θ = e i θ \text{cis } \theta = cos \theta + i \sin \theta = e^{i\theta} . Therefore,

∑ k = 1 15 cis 2 k − 1 π 36 = ∑ k = 1 15 e i ( 2 k − 1 ) π 36 This is a GP with first term e i π 36 and common ratio e i π 18 = e i π 36 ( 1 − e i 15 π 18 ) 1 − e i π 18 = ( cos ⁡ π 36 + i sin ⁡ π 36 ) ( 1 − cos ⁡ 5 π 6 − i sin ⁡ 5 π 6 ) 1 − cos ⁡ π 18 − i sin ⁡ π 18 = ( cos ⁡ π 36 + i sin ⁡ π 36 ) ( 1 − cos ⁡ 5 π 6 − i sin ⁡ 5 π 6 ) ( 1 − cos ⁡ π 18 + i sin ⁡ π 18 ) ( 1 − cos ⁡ π 18 − i sin ⁡ π 18 ) ( 1 − cos ⁡ π 18 + i sin ⁡ π 18 ) = ( cos ⁡ π 36 + i sin ⁡ π 36 ) ( 1 + 3 2 − i 1 2 ) ( 1 − ( 1 − 2 sin ⁡ 2 π 36 ) + i 2 sin ⁡ π 36 cos ⁡ π 36 ) ( 1 − cos ⁡ π 18 ) 2 + sin ⁡ 2 π 18 = ( cos ⁡ π 36 + i sin ⁡ π 36 ) ( 2 + 3 − i ) sin ⁡ π 36 ( sin ⁡ π 36 + i cos ⁡ π 36 ) 2 − 2 cos ⁡ π 18 = sin ⁡ π 36 ( 2 + 3 − i ) ( cos ⁡ π 36 sin ⁡ π 36 − sin ⁡ π 36 cos ⁡ π 36 + i ( sin ⁡ 2 π 36 + cos ⁡ 2 π 36 ) ) 4 sin ⁡ 2 π 36 = 1 + i ( 2 + 3 ) 4 sin ⁡ π 36 \begin{aligned} \sum_{k=1}^{15} \text{cis }^{2k-1} \frac{\pi}{36} & = \sum_{k=1}^{15} e^{i\frac{(2k-1)\pi}{36}} \quad \quad \small \color{#3D99F6}{\text{This is a GP with first term }e^{i\frac{\pi}{36}} \text{ and common ratio }e^{i\frac{\pi}{18}}} \\ & = \frac{e^{i\frac{\pi}{36}}\left(1 - e^{i\frac{15\pi}{18}}\right)}{1- e^{i\frac{\pi}{18}}} \\ & = \frac{\left(\cos \frac{\pi}{36} + i \sin \frac{\pi}{36}\right) \left(1 - \cos \frac{5\pi}{6} - i \sin \frac{5\pi}{6} \right)}{1 - \cos \frac{\pi}{18} - i \sin \frac{\pi}{18}} \\ & = \frac{\left(\cos \frac{\pi}{36} + i \sin \frac{\pi}{36}\right) \left(1 - \cos \frac{5\pi}{6} - i \sin \frac{5\pi}{6} \right)\left(1 - \cos \frac{\pi}{18} + i \sin \frac{\pi}{18}\right)}{\left(1 - \cos \frac{\pi}{18} - i \sin \frac{\pi}{18} \right) \left(1 - \cos \frac{\pi}{18} + i \sin \frac{\pi}{18}\right)} \\ & = \frac{\left(\cos \frac{\pi}{36} + i \sin \frac{\pi}{36}\right) \left(1 + \frac{\sqrt{3}}{2} - i \frac{1}{2} \right)\left(1 - (1 - 2 \sin^2 \frac{\pi}{36}) + i 2 \sin \frac{\pi}{36} \cos \frac{\pi}{36} \right)}{\left(1 - \cos \frac{\pi}{18} \right)^2 + \sin^2 \frac{\pi}{18}} \\ & = \frac {\left(\cos \frac{\pi}{36} + i \sin \frac{\pi}{36}\right) \left(2 + \sqrt{3} - i \right) \sin \frac{\pi}{36} \left(\sin \frac{\pi}{36} + i \cos \frac{\pi}{36}\right)}{2 - 2 \cos \frac{\pi}{18}} \\ & = \frac { \sin \frac{\pi}{36} \left(2 + \sqrt{3} - i \right) \left(\cos \frac{\pi}{36} \sin \frac{\pi}{36} - \sin \frac{\pi}{36} \cos \frac{\pi}{36} + i(\sin^2 \frac{\pi}{36} + \cos^2 \frac{\pi}{36}) \right)}{4 \sin^2 \frac{\pi}{36}} \\ & = \frac{1+i(2+\sqrt{3})}{4 \sin \frac{\pi}{36}} \end{aligned}

⇒ ℑ ( ∑ k = 1 15 cis 2 k − 1 π 36 ) = 2 + 3 4 sin ⁡ π 36 \begin{aligned} \Rightarrow \Im \left( \sum_{k=1}^{15} \text{cis }^{2k-1} \frac{\pi}{36}\right) & = \boxed{\dfrac{2+\sqrt{3}}{4 \sin \frac{\pi}{36}}} \end{aligned}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Mr Cheong, you really are the trigonometry god!!!

Jun Arro Estrella - 5 years, 5 months ago

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Not really. You may Brian Charlesworth better. I am glad that you like the solution.

Chew-Seong Cheong - 5 years, 5 months ago

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I really like how you lay down your solutions Mr Cheong.. you are a living legend and I am grateful to have lived and see it :) Keep it up..

Jun Arro Estrella - 5 years, 5 months ago

Let w = e π i / 36 w=e^{\pi i/36} . Then let S = ∑ k = 1 15 w 2 k − 1 S=\displaystyle \sum_{k=1}^{15} w^{2k-1} . Using the formula for the sum of a geometric progression we have:

S = w − 1 ∑ k = 1 15 w 2 k = w − 1 ⋅ w 32 − w 2 w 2 − 1 = w 31 − w w 2 − 1 S=w^{-1} \displaystyle \sum_{k=1}^{15} w^{2k}=w^{-1}\cdot\dfrac{w^{32}-w^2}{w^2-1}=\dfrac{w^{31}-w}{w^2-1}

Now, simplify and rationalize the denominator:

S = ( w 31 − w ) ( w − 2 − 1 ) ( w 2 − 1 ) ( w − 2 − 1 ) = w 29 − w 31 − w − 1 + w 1 − w 2 − w − 2 + 1 = w 29 − w 31 − w − 1 + w 2 − 2 cos ⁡ ( π 18 ) S=\dfrac{(w^{31}-w)(w^{-2}-1)}{(w^2-1)(w^{-2}-1)}=\dfrac{w^{29}-w^{31}-w^{-1}+w}{1-w^2-w^{-2}+1}=\dfrac{w^{29}-w^{31}-w^{-1}+w}{2-2\cos\left(\dfrac{\pi}{18}\right)}

Finally, find the real part and the imaginary part, and simplify the denominator using the identity sin ⁡ ( θ 2 ) = 1 − cos ⁡ θ 2 \sin\left(\dfrac{\theta}{2}\right)=\sqrt{\dfrac{1-\cos \theta}{2}} :

S = cos ⁡ ( 29 π 36 ) + i sin ⁡ ( 29 π 36 ) − cos ⁡ ( 31 π 36 ) − i sin ⁡ ( 31 π 36 ) − cos ⁡ ( π 36 ) + i sin ⁡ ( π 36 ) + cos ⁡ ( π 36 ) + i sin ⁡ ( π 36 ) 4 sin ⁡ 2 ( π 36 ) S=\dfrac{\cos\left(\dfrac{29\pi}{36}\right)+i\sin\left(\dfrac{29\pi}{36}\right)-\cos\left(\dfrac{31\pi}{36}\right)-i\sin\left(\dfrac{31\pi}{36}\right)-\cos\left(\dfrac{\pi}{36}\right)+i\sin\left(\dfrac{\pi}{36}\right)+\cos\left(\dfrac{\pi}{36}\right)+i\sin\left(\dfrac{\pi}{36}\right)}{4\sin^2\left(\dfrac{\pi}{36}\right)}

S = cos ⁡ ( 29 π 36 ) − cos ⁡ ( 31 π 36 ) + i sin ⁡ ( 29 π 36 ) − i sin ⁡ ( 31 π 36 ) + 2 i sin ⁡ ( π 36 ) 4 sin ⁡ 2 ( π 36 ) S=\dfrac{\cos\left(\dfrac{29\pi}{36}\right)-\cos\left(\dfrac{31\pi}{36}\right)+i\sin\left(\dfrac{29\pi}{36}\right)-i\sin\left(\dfrac{31\pi}{36}\right)+2i\sin\left(\dfrac{\pi}{36}\right)}{4\sin^2\left(\dfrac{\pi}{36}\right)}

So, the imaginary part is:

Im ( S ) = sin ⁡ ( 29 π 36 ) − sin ⁡ ( 31 π 36 ) + 2 sin ⁡ ( π 36 ) 4 sin ⁡ 2 ( π 36 ) \text{Im}(S)=\dfrac{\sin\left(\dfrac{29\pi}{36}\right)-\sin\left(\dfrac{31\pi}{36}\right)+2\sin\left(\dfrac{\pi}{36}\right)}{4\sin^2\left(\dfrac{\pi}{36}\right)}

Use the sum to product formulas:

Im ( S ) = 2 cos ⁡ ( 5 π 6 ) sin ⁡ ( − π 36 ) + 2 sin ⁡ ( π 36 ) 4 sin ⁡ 2 ( π 36 ) \text{Im}(S)=\dfrac{2\cos\left(\dfrac{5\pi}{6}\right)\sin\left(-\dfrac{\pi}{36}\right)+2\sin\left(\dfrac{\pi}{36}\right)}{4\sin^2\left(\dfrac{\pi}{36}\right)}

Im ( S ) = 2 + 3 4 sin ⁡ ( π 36 ) \text{Im}(S)=\dfrac{2+\sqrt{3}}{4\sin\left(\dfrac{\pi}{36}\right)}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

did it a similar way too long enough sigh.

Mardokay Mosazghi - 5 years, 5 months ago

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