Answers

2016-02-06T20:02:33+05:30
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

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