Question #9c7b3

Nov 25, 2016

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

Explanation:

If you know the de Moivre's identity it is quite direct.

${e}^{i x} = \cos x + i \sin x$

so

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

or

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

grouping the real and the imaginary parts,

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

now, using the identity

${\cos}^{2} x + {\sin}^{2} x = 1$ we will get at

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

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

Nov 25, 2016

Expand ${\sin}^{3} x$

Explanation:

There is another simpler method.
Use the trig identity:
$\sin 3 x = 3 \sin x - 4 {\sin}^{3} x$
By transposing terms, we get:
$4 {\sin}^{3} x = 3 \sin x - \sin 3 x$
${\sin}^{3} x = \frac{3 \sin x - \sin 3 x}{4}$