sin pie/7 * sin 2pie/7 * sin 3pie/7 = sq.root7 / 8

Asked by Satyam Sharma | 15th Apr, 2013, 06:23: PM

Expert Answer:

Now, 

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

Answered by  | 16th Apr, 2013, 06:32: AM

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