How do you verify the identity $sin4x= 4sinxcosx(1-sin^2x)$ ?

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Answer 1 · 2015-04-23T20:08:31

Remember
$sin(2theta) = 2*sin(theta)*cos(theta)$
and
$cos(2theta) = 1 - 2*sin^2(theta)$

So
$sin(4x)$

$= 2*sin(2x)*cos(2x)$

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

$=4sin(x)cos(x)(1-sin^2(x))$

How do you verify the identity sin4x= 4sinxcosx(1-sin^2x)sin4x= 4sinxcosx(1-sin^2x)?