Prove : ((Sin 2A ) / (2 sin 3A) )(4(cos 2A)^2 -1) = cos 3A ?

1 Answer
May 15, 2017

See proof below

Explanation:

We need

sin^2A+cos^2A=1

cos2A=2cos^2A-1

sin2A=2sinAcosA

sin3A=sin2AcosA+cos2AsinA

=2sinAcos^2A+sinA(2cos^2A-1)

=sinA(-1+4cos^2A)

cos3A=cos2AcosA-sin2AsinA

=(2cos^2A-1)cosA-2sinAcosAsinA

=2cos^3A-cosA-2cosA(1-cos^2A)

=4cos^3A-3cosA

Therefore,

LHS=(sin2A)/(2sin3A)*(4(cos2A)^2-1)

=(cancel(2sinA)cosA)/(cancel(2sinA)(-1+4cos^2A))*(4(2cos^2A-1)^2-1)

=cosA/((-1+4cos^2A))*(16cos^4A-16cos^2A+4-1)

=cosA/(cancel((4cos^2A-1)))*cancel((4cos^2A-1))(4cos^2A-3)

=cosA(4cos^2A-3)

=4cos^3A-3cosA

=cos3A

=RHS

QED