Question #f676d

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Question #f676d

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Answer 1 · 2016-10-19T07:26:13

We will use the double angle identities

$sin (2 θ) = 2 sin (θ) cos (θ)$ $sin(2theta) = 2sin(theta)cos(theta)$
$cos (2 θ) = 1 - 2 {sin}^{2} (θ) = {cos}^{2} (θ) - {sin}^{2} (θ) = 2 {cos}^{2} (θ) - 1$ $cos(2theta) = 1-2sin^2(theta) = cos^2(theta)-sin^2(theta) = 2cos^2(theta)-1$

Proceeding...

$\frac{1}{2} sin (4 x) = \frac{1}{2} sin (2 (2 x))$ $1/2sin(4x) = 1/2sin(2(2x))$

$= \frac{1}{2} (2 sin (2 x) cos (2 x))$ $= 1/2(2sin(2x)cos(2x))$

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

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

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

$= 2 sin (x) cos (x) - 4 {sin}^{3} (x) cos (x)$ $=2sin(x)cos(x) - 4sin^3(x)cos(x)$

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Question #f676d

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