CBSE Class 11-science Answered

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

Now, since sin(pi+x) = -sinx = -sin(pi-x)

hence, sin2pi/7 + sin4pi/7 + sin (pi-8pi/7) = [(7)^1/2]/2

hence, sin2pi/7 + sin4pi/7 - sin (pi/7) = [(7)^1/2]/2

sin2pi/7 + sin4pi/7 - sin (pi-pi/7) = [(7)^1/2]/2

sin2pi/7 + sin4pi/7 - sin (6pi/7) = [(7)^1/2]/2

Substituting this back in 1

= 1/4*[(7)^1/2]/2

= [(7)^1/2]/8

Hence proved