How do you express cot^3theta+tan^3theta cot3θ+tan3θ in terms of non-exponential trigonometric functions?

1 Answer
Oct 15, 2016

cot^3theta+tan^3theta cot3θ+tan3θ

=cos^3theta/sin^3theta+sin^3theta/cos^3theta=cos3θsin3θ+sin3θcos3θ

=(cos^6theta+sin^6theta)/(sin^3thetacos^3theta)=cos6θ+sin6θsin3θcos3θ

=((cos^2theta+sin^2theta)^3-3(cos^2thetasin^2theta)(cos^2theta+sin^2theta))/(sin^3thetacos^3theta)=(cos2θ+sin2θ)33(cos2θsin2θ)(cos2θ+sin2θ)sin3θcos3θ

=(1-3(cos^2thetasin^2theta)*1)/(sin^3thetacos^3theta)=13(cos2θsin2θ)1sin3θcos3θ

=(8-24(cos^2thetasin^2theta))/(2^3sin^3thetacos^3theta)=824(cos2θsin2θ)23sin3θcos3θ

=(8-6(2costhetasintheta)^2)/(2sinthetacostheta)^3=86(2cosθsinθ)2(2sinθcosθ)3

=(8-3*2sin^2 2theta)/(sin^3 2theta)=832sin22θsin32θ

=(8-3*(1-cos4theta))/(1/4(3sin2theta-sin6theta)=83(1cos4θ)14(3sin2θsin6θ)

=(5+3cos4theta)/(1/4(3sin2theta-sin6theta)=5+3cos4θ14(3sin2θsin6θ)

=(20+12cos4theta)/(3sin2theta-sin6theta)=20+12cos4θ3sin2θsin6θ