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Question #82719

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May 23, 2016

We will use the following identities:

$sin (α + β) = sin (α) cos (β) + cos (α) sin (β)$ $sin(alpha + beta) = sin(alpha)cos(beta)+cos(alpha)sin(beta)$
$sin (2 x) = 2 sin (x) cos (x)$ $sin(2x) = 2sin(x)cos(x)$
$cos (2 x) = 2 {cos}^{2} (x) - 1$ $cos(2x) = 2cos^2(x)-1$

With those, we have

$sin (3 x) = sin (x + 2 x)$ $sin(3x) = sin(x + 2x)$

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

$= sin (x) (2 {cos}^{2} (x) - 1) + cos (x) (2 sin (x) cos (x))$ $=sin(x)(2cos^2(x)-1)+cos(x)(2sin(x)cos(x))$

$= 2 {cos}^{2} (x) sin (x) - sin (x) + 2 {cos}^{2} (x) sin (x)$ $=2cos^2(x)sin(x)-sin(x)+2cos^2(x)sin(x)$

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

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