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

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Answer 1 · 2017-05-15T10:24:00

See proof below

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#