Question

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

Answer 1

`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