Show that:

sin(2*pi/7) + sin(4*pi/7) + sin(8*pi/7) = √7/2

Note: All the angles are in radians.

Hi!

So let's start from z^7 = 1.

This equation, in complex have 7 solution, like this

z(k) = cos(2*k*pi/7)+i*sin(2*k*pi/7) with k=0..6

From Viete we know that the sum of all solution are 0 

So we find that :

1+cos(2*pi/7)+cos(4*pi/7)+cos(6*pi/7)+cos(8*pi/7)

+cos(10*pi/7)+cos(12*pi/7)=0.

After i transform some trigonometric function i find a nice relation

cos(pi/7)-cos(2*pi/7)+cos(3*pi/7) = 1/2  (relation 1)

Let be x=sin(2*pi/7)+sin(4*pi/7)+sin(8*pi/7). 

After new transformation i find that

x=-sin(pi/7)+sin(2*pi/7)+sin(3*pi/7). (relation 2)

I square both relation (1) and (2) and i find :

x^2 + 1/4 = 3-2(cos(pi/7)-cos(2*pi/7)+cos(3*pi/7)). 

So x^2+1/4 = 3-2/2=3-1=2

So x^2=2-1/4=7/4

And x=sqrt(7)/2.

A really nice problem!

Edited on November 15, 2007, 4:13 pm

Posted by Chesca Ciprian on 2007-11-15 16:11:31