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

1 Answer
Jan 15, 2016

Start with sin(2theta)=2costhetasintheta ...

Explanation:

sin(2theta)=2costhetasintheta

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

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

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

(sin^3(2theta))/8=(1/8)[(sin2theta)(sin2theta)]xxsin2theta

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

=(1/8)[1/2(1-cos4theta)]xxsin2theta

=(1/16)(1-4costheta)(sin2theta)

Hope that helped