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

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