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aditya prakash morey Grade: 11
        prove that: sin2pi/7 + sin4pi/7 + sin8pi/7  = (root7)/2
7 years ago

Answers : (1)

Ramesh V
70 Points
										

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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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

7 years ago
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