How do you express sin^4theta - sin^3theta *cos^2theta in terms of non-exponential trigonometric functions?

1 Answer
P dilip_k
Apr 30, 2016

sin^4theta-sin^3theta*cos^2theta

=sin^3theta(sintheta-cos^2theta)

now
sin3theta =3sintheta -4sin^3theta=>sin^3theta=3/4sintheta-1/4sin3theta
and cos^2theta=1/2(1+cos2theta)

Putting these in the main expression it becomes
=sin^3theta(sintheta-cos^2theta)
=(3/4sintheta-1/4sin3theta)(sintheta-1/2(1+cos2theta))
=(3/4sin^2theta-1/4sin3thetasintheta-3/8sintheta-3/8sinthetacos2theta+1/8sin3theta-1/4sin3theta cos2theta)

=(3/8(1-cos2theta)-1/4sin3thetasintheta-3/8sintheta-3/8sinthetacos2theta+1/8sin3theta-1/4sin3theta cos2theta)
=(3/8(1-cos2theta)-1/8(cos2theta-cos4theta)-3/8sintheta-3/16(sin3theta-sintheta)+1/8sin3theta-1/8(sin5theta-sintheta))

Pl proceed for simlification

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