How do you prove $9 sin (3 x) cos (3 x)$ $9sin(3x)cos(3x)$ ?

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Answer 1 · 2016-02-05T23:11:05

$(\frac{9}{2}) sin (6 x)$ $(9/2)sin(6x)$

we know that

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

then:

$9 sin (3 x) cos (3 x) = (\frac{9}{2}) \cdot 2 sin (3 x) cos (3 x)$ $9sin(3x)cos(3x)=(9/2)*2sin(3x)cos(3x)$

$(\frac{9}{2}) sin (6 x)$ $(9/2)sin(6x)$

How do you prove 9sin(3x)cos(3x)9sin(3x)cos(3x)9sin(3x)cos(3x)?