What is cos^3thetasin^3thetacos3θsin3θ in terms of non-exponential trigonometric functions?

1 Answer
Jan 15, 2016

Start with sin(2theta)=2costhetasinthetasin(2θ)=2cosθsinθ ...

Explanation:

sin(2theta)=2costhetasinthetasin(2θ)=2cosθsinθ

Now divide both sides by 2, then cube it ...

[(sin(2theta))/2]^3=[costhetasintheta]^3=cos^3thetasin^3theta[sin(2θ)2]3=[cosθsinθ]3=cos3θsin3θ

So, now we have the original equation equal to (sin^3(2theta))/8sin3(2θ)8

(sin^3(2theta))/8=(1/8)[(sin2theta)(sin2theta)]xxsin2thetasin3(2θ)8=(18)[(sin2θ)(sin2θ)]×sin2θ

Now use the sinthetasinthetasinθsinθ product-to-sum identity ...

=(1/8)[1/2(1-cos4theta)]xxsin2theta=(18)[12(1cos4θ)]×sin2θ

=(1/16)(1-4costheta)(sin2theta)=(116)(14cosθ)(sin2θ)

Hope that helped