How do you prove 8sin^2xcos^3x = cosx - cosxcos(4x)?

To show that
#8sin^2thetacos^3theta=costheta-costhetacos(4theta)#

#8sin^2thetacos^3theta=costheta(1-cos4theta)#

Dividing by costheta

#8sin^2thetacos^2theta=1-cos4theta#

#1-cos4theta=2sin^2(2theta)#

#8sin^2thetacos^2theta=2sin^2(2theta)#

#2sin^2(2theta)=2(sin2theta)^2#

#sin2theta=2sinthetacostheta#

#2(sin2theta)^2=2(2sinthetacostheta)^2#

#2sin^2(2theta)=2xx4sin^2thetacos^2theta#

#2sin^2(2theta)=8sin^2thetacos^2theta#

#1-cos4theta=2sin^2(2theta)#

#1-cos4theta=8sin^2thetacos^2theta#

#costheta(1-cos4theta)=costheta(8sin^2thetacos^2theta)#

Interchanging

#8sin^2thetacos^3theta=costheta-costhetacos(4theta)#

#LHS=8sin^2thetacos^3theta=8sin^2thetacos^2thetacostheta=2costheta(2sinthetacostheta)^2=2costheta(sin2theta)^2=costheta(2sin^2(2theta))=costheta(1-cos4theta)=costheta-costhetacos4theta=RHS#
Note:
1. #sin2theta=2sinthetacostheta#
2. #sin^2(2theta)=(1-cos4theta)/2#.