prove that: sin2pi/7 + sin4pi/7 + sin8pi/7 = (root7)/2

Lets take a = 2pi/7

7a =2pi
sin4a = sin(2pi-3a)
sin4a  = -sin3a
2sin2a.cos2a = 4sin³(a) -3sina
4sin a.cos a (1-2sin²(a)) = sin a(4sin²(a) - 3)
4cos a(1-2sin²(a)) = 4sin²(a) -3
On squaring both sides

16(1-sin²(a)) [1-2sin²(a))]^2 = (4sin²(a) -3)²
64sin⁶(a) - 112sin⁴(a) + 56sin²(a) -7 =0
it is cubic in sin²(a)
its roots are sin²(2pi/7) ,sin⁴(pi/7) ,sin²(8pi/7)
sum of roots =7/4
sin2pi/7*sin4pi/7 +sin4pi/7*sin8pi/7 +sin8pi/7*sin2pi/7 = 0
we can simply prove it by using 2sin a.sin b= cos(a-b) - cos(a+b)
& cos(2pi-theta) = cos theta

(sin2pi/7+sin4pi/7 +sin8pi/7)²=7/4
sin2pi/7+sin4pi/7 +sin8pi/7=[(7)^1/2]/2

--

Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and
we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Naga Ramesh
IIT Kgp - 2005 batch