What is ${cos}^{3} θ {sin}^{3} θ$ $cos^3thetasin^3theta$ in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-01-15T01:15:21

Start with $sin (2 θ) = 2 cos θ sin θ$ $sin(2theta)=2costhetasintheta$ ...

$sin (2 θ) = 2 cos θ sin θ$ $sin(2theta)=2costhetasintheta$

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

${[\frac{sin (2 θ)}{2}]}^{3} = {[cos θ sin θ]}^{3} = {cos}^{3} θ {sin}^{3} θ$ $[(sin(2theta))/2]^3=[costhetasintheta]^3=cos^3thetasin^3theta$

So, now we have the original equation equal to $\frac{{sin}^{3} (2 θ)}{8}$ $(sin^3(2theta))/8$

$\frac{{sin}^{3} (2 θ)}{8} = (\frac{1}{8}) [(sin 2 θ) (sin 2 θ)] \times sin 2 θ$ $(sin^3(2theta))/8=(1/8)[(sin2theta)(sin2theta)]xxsin2theta$

Now use the $sin θ sin θ$ $sinthetasintheta$ product-to-sum identity ...

$= (\frac{1}{8}) [\frac{1}{2} (1 - cos 4 θ)] \times sin 2 θ$ $=(1/8)[1/2(1-cos4theta)]xxsin2theta$

$= (\frac{1}{16}) (1 - 4 cos θ) (sin 2 θ)$ $=(1/16)(1-4costheta)(sin2theta)$

Hope that helped

What is cos^3thetasin^3thetacos3θsin3θcos^3thetasin^3theta in terms of non-exponential trigonometric functions?