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# How do you prove sin3x = 3sinx - 4sin^3x?

Apr 26, 2015

Since the angle sum formula of sine is:

$\sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$,

and the double angle formula of cosine:

$\cos \left(2 \alpha\right) = {\cos}^{2} \alpha - {\sin}^{2} \alpha = 2 {\cos}^{2} \alpha - 1 = 1 - 2 {\sin}^{2} \alpha$

then:

$\sin \left(3 x\right) = \sin \left(2 x + x\right) = \sin \left(2 x\right) \cos x + \cos \left(2 x\right) \sin x =$

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

$= 2 \sin x {\cos}^{2} x + \sin x - 2 {\sin}^{3} x =$

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

$= 2 \sin x - 2 {\sin}^{3} x + \sin x - 2 {\sin}^{3} x =$

$= 3 \sin x - 4 {\sin}^{3} x$.

Mar 23, 2018

we can use DeMoivre's theorem

#### Explanation:

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta + i \sin n \theta - - \left(1\right)$

since we want ${\sin}^{3} x$ we will expand ${\left(\cos x + i \sin x\right)}^{3}$

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

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

from$\left(1\right)$

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

equating imaginary parts

$\sin 3 x = 3 {\cos}^{2} x \sin x - {\sin}^{3} x$

but ${\cos}^{2} x = 1 - {\sin}^{2} x$

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

$\sin 3 x = 3 \sin x - 3 {\sin}^{2} x \sin x - {\sin}^{3} x$

$\sin 3 x = 3 \sin x - 4 {\sin}^{3} x$

as required