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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