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

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Answer 1 · 2016-01-15T01:15:21

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

#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