How do you solve $cos^3xsin^2x=(sin^2x-sin^4x)cosx$ ?

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Answer 1 · 2017-06-25T12:47:56

$LHS=cos^3xsin^2x$

$=cosxcos^2xsin^2x$

$=cosx(1-sin^2x)sin^2x$

$=cosx(sin^2x-sin^4x)=RHS$

How do you solve cos^3xsin^2x=(sin^2x-sin^4x)cosxcos^3xsin^2x=(sin^2x-sin^4x)cosx?