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# Prove that cos2?/7+cos4?/7+cos6?/7=-1/2.

Asked by SIVENDU SIVENDU 28th July 2012, 5:42 PM

Answer : To prove : cos(2pi/7)+cos(4pi/7)+cos(6pi/7) = - 1 / 2.

Multiplying both sides by sin (pi /7)

=> sin(pi /7)cos(2pi /7)+sin(pi /7)cos(4pi /7)+sin(pi /7)cos(6pi /7) = -sin(pi/7)/2

By transform the product in sum:

=> (1/2)[sin(pi /7-2pi /7) + sin(pi /7+2pi /7)] + (1/2)[sin(pi /7-4pi /7) + sin(pi/7+4pi /7)] + (1/2)[sin(pi /7-6pi /7) + sin(pi /7+6pi /7)]= -sin(pi /7)/2

Now  dividing by (1/2) both sides , we get

=> -sin(pi /7) + sin(3pi /7) - sin(3pi /7) + sin(5pi /7) - sin(5pi /7) +  sin(7pi /7) = -sin(pi /7)

Eliminating like terms , we get

=> -sin(pi /7) + sin(pi )= -sin(pi /7)

But sin(pi) = 0

-sin(pi /7) = -sin(pi /7)

LHS = RHS

Hence Proved

Answered by Expert 28th July 2012, 6:22 PM
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