How do you express sin^4theta - sin^3theta *cos^2thetasin4θsin3θcos2θ in terms of non-exponential trigonometric functions?

1 Answer
Apr 30, 2016

sin^4theta-sin^3theta*cos^2thetasin4θsin3θcos2θ

=sin^3theta(sintheta-cos^2theta)=sin3θ(sinθcos2θ)

now
sin3theta =3sintheta -4sin^3theta=>sin^3theta=3/4sintheta-1/4sin3thetasin3θ=3sinθ4sin3θsin3θ=34sinθ14sin3θ
and cos^2theta=1/2(1+cos2theta)cos2θ=12(1+cos2θ)

Putting these in the main expression it becomes
=sin^3theta(sintheta-cos^2theta)=sin3θ(sinθcos2θ)
=(3/4sintheta-1/4sin3theta)(sintheta-1/2(1+cos2theta))=(34sinθ14sin3θ)(sinθ12(1+cos2θ))
=(3/4sin^2theta-1/4sin3thetasintheta-3/8sintheta-3/8sinthetacos2theta+1/8sin3theta-1/4sin3theta cos2theta)=(34sin2θ14sin3θsinθ38sinθ38sinθcos2θ+18sin3θ14sin3θcos2θ)

=(3/8(1-cos2theta)-1/4sin3thetasintheta-3/8sintheta-3/8sinthetacos2theta+1/8sin3theta-1/4sin3theta cos2theta)=(38(1cos2θ)14sin3θsinθ38sinθ38sinθcos2θ+18sin3θ14sin3θcos2θ)
=(3/8(1-cos2theta)-1/8(cos2theta-cos4theta)-3/8sintheta-3/16(sin3theta-sintheta)+1/8sin3theta-1/8(sin5theta-sintheta))=(38(1cos2θ)18(cos2θcos4θ)38sinθ316(sin3θsinθ)+18sin3θ18(sin5θsinθ))

Pl proceed for simlification