# Show that sin 3x = 3 sin x - 4sin^3 x ?

Aug 27, 2015

$\sin \left(A \pm B\right) = \sin A \cos B \pm \cos A \sin B$
$\sin 2 A = 2 \sin A \cos A$
$\cos 2 A = {\cos}^{2} A - {\sin}^{2} A$
${\sin}^{2} A + {\cos}^{2} A = 1$

$\sin 3 x = \sin x \cos 2 x + \cos x \sin 2 x = \sin x \left({\cos}^{2} x - {\sin}^{2} x\right) + 2 \sin x {\cos}^{2} x = 3 \sin x {\cos}^{2} x - {\sin}^{3} x = 3 \sin x \left(1 - {\sin}^{2} x\right) - {\sin}^{3} x = 3 \sin x - 4 {\sin}^{3} x$

Aug 27, 2015

Alternatively, use Euler's Formula, expand $\cos \left(3 x\right) + i \sin \left(3 x\right)$ and equate the imaginary parts.

#### Explanation:

If you have encountered Euler's Formula then it works out like this:

Euler's Formula gives us: $\cos x + i \sin x = {e}^{i x}$

So:

$\cos \left(3 x\right) + i \sin \left(3 x\right) = {e}^{i 3 x} = {\left({e}^{i x}\right)}^{3} = {\left(\cos \left(x\right) + i \sin \left(x\right)\right)}^{3}$

$= {\cos}^{3} \left(x\right) + 3 {\cos}^{2} \left(x\right) i \sin \left(x\right) + 3 \cos \left(x\right) {i}^{2} {\sin}^{2} \left(x\right) + {i}^{3} {\sin}^{3} \left(x\right)$

$= \left({\cos}^{3} \left(x\right) - 3 \cos \left(x\right) {\sin}^{2} \left(x\right)\right) + i \left(3 {\cos}^{2} \left(x\right) \sin \left(x\right) - {\sin}^{3} \left(x\right)\right)$

Equating the imaginary parts of both ends, we get:

$\sin \left(3 x\right) = 3 {\cos}^{2} \left(x\right) \sin \left(x\right) - {\sin}^{3} \left(x\right)$

$= \sin \left(x\right) \left(3 {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)\right)$

$= \sin \left(x\right) \left(3 \left(1 - {\sin}^{2} \left(x\right)\right) - {\sin}^{2} \left(x\right)\right)$ $\textcolor{g r e e n}{\left[\left[\text{using: } {\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1\right]\right]}$

$= \sin \left(x\right) \left(3 - 4 {\sin}^{2} \left(x\right)\right)$

$= 3 \sin \left(x\right) - 4 {\sin}^{3} \left(x\right)$

Equating the real parts of both ends, we also get:

$\cos \left(3 x\right) = {\cos}^{3} \left(x\right) - 3 \cos \left(x\right) {\sin}^{2} \left(x\right)$

$= \cos \left(x\right) \left({\cos}^{2} \left(x\right) - 3 {\sin}^{2} \left(x\right)\right)$

$= \cos \left(x\right) \left({\cos}^{2} \left(x\right) - 3 \left(1 - {\cos}^{2} \left(x\right)\right)\right)$

$= \cos \left(x\right) \left(4 {\cos}^{2} \left(x\right) - 3\right)$

$= 4 {\cos}^{3} \left(x\right) - 3 \cos \left(x\right)$