How do you verify the identity of: $cos^3x sin^2x=(sin^2x-sin^4x)cosx$ ?

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Answer 1 · 2015-04-22T17:10:51

From basic definitions and the Pythagorean Theorem
$cos^2(x)+sin^2(x)=1$
or
$cos^2(x) = 1-sin^2(x)$

First consider
$(sin^2(x)-sin^4(x)$

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

$=underbrace(sin^2(x)cos^2(x))_(used below)$

So
$cos^3(x)sin^2(x)$

$=(cos(x)) * [underbrace((cos^2(x)sin^2(x)))_(as above)]$

$= (cos(x))*(sin^2(x)-sin^4(x))$

$= (sin^2(x)-sin^4(x))cos(x)$

How do you verify the identity of: cos^3x sin^2x=(sin^2x-sin^4x)cosxcos^3x sin^2x=(sin^2x-sin^4x)cosx?