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Prove $$ \sin(2\pi/7) +\sin(4\pi/7) +\sin(8\pi/7) = \sqrt7/2$$

I have squared both sides and have come to the stage where the LHS is as follows

$$3/2 -1/2 (\cos{4\pi/7} + \cos{8\pi/7} + \cos{16\pi/7})$$

How do I get it to equal 7/4

PS; I DO NOT KNOW ANYTHING ABOUT COMPLEX NUMBERS

sidt36
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1 Answers1

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Let put $7\theta$=$2n\pi$ where $n$ stands for any integer. Now $$7\theta=2n\pi$$$$4\theta=2n\pi-3\theta$$$$\Rightarrow \sin(4\theta)=\sin(2n\pi-3\theta)$$ $$\rightarrow \sin(4\theta)=-\sin(3\theta)\tag1$$

Now from equation $1$ we can get

$$2\sin(2\theta)\cos(2\theta)=4\sin^3\theta-3\sin\theta$$ $$4\sin\theta \cos\theta (1-2\sin^2\theta)=\sin\theta(4\sin^2\theta-3)$$ $$4\cos\theta(1-2\sin^2\theta)=4\sin^2\theta-3$$ Squaring both sides and then adding the result, you will get $$64\sin^6\theta-112\sin^4\theta+56\sin^2\theta-7=0$$

This is a cubic with $\sin^2\theta$ with the roots:$\sin^2(2\pi /7)$, $\sin^2(4\pi /7)$, $\sin^2(8\pi /7)$

Sum of the roots is $$\sin^2(2\pi /7)+\sin^2(4\pi /7)+\sin^2(8\pi /7)=\frac{7}{4}\tag2$$

But you can prove through trigonometric identities (as sum of all the angles is $2\pi$) that $$\sin(2\pi /7) \sin(4\pi /7)+\sin(4\pi /7) \sin(8\pi /7)+\sin(2\pi /7)\sin(8\pi /7)=0$$

Hence the sum of the roots from $(2)$ $$(\sin(2\pi /7)+\sin(4\pi /7)+\sin(8\pi /7))^2=\frac{7}{4}$$ And the result follows.

egreg
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Harsh Sharma
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    How do you come to the conclusion that the roots are$$sin^2(2\pi /7),sin^2(4\pi /7),sin^2(8\pi /7)$$ – sidt36 Jul 04 '16 at 14:34
  • @sidt36 Because $\theta=2n\pi /7$ – Harsh Sharma Jul 04 '16 at 14:50
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    Ah....Finally I get it...But Why can't it be $\sin^2 10\pi/7$ or any other even multiple of $\pi/7$ – sidt36 Jul 04 '16 at 14:55
  • @sidt36 Ya you are right it can be. For example check out for $20\pi /7, 10\pi /7$, it would surely give the result. I just use the starting three to get my work done. – Harsh Sharma Jul 04 '16 at 15:03
  • @Harsh Sharma "But you can prove through trigonometric identities (as sum of all the angles is $2\pi$) that $$\sin(2\pi /7) \sin(4\pi /7)+\sin(4\pi /7) \sin(8\pi /7)+\sin(2\pi /7)\sin(8\pi /7)=0$$" how can u show it ? I did differently and it was lengthy –  Mar 23 '21 at 07:42