A Formula for sin(3x)

The prupose of this page is to prove the following formula:

$\sin 3x =4\sin x\sin(60^{\circ}-x)\sin(60^{\circ}+x).$

We first remind of another useful trigonometric identity:

$\displaystyle\sin\alpha + \sin\beta +\sin\gamma -\sin(\alpha +\beta +\gamma)=4\sin\frac{\alpha +\beta}{2}\sin\frac{\beta +\gamma}{2}\sin\frac{\gamma +\alpha}{2}.$

Taking here $\alpha =x,$ $\beta =x+120^{\circ},$ $\gamma =x-120^{\circ}$ reduces the formula to

$\sin x + \sin (x+120^{\circ}) +\sin(x-120^{\circ}) -\sin 3x=4\sin x \sin (x+60^{\circ})\sin (x-60^{\circ}),$

implying that in order to establish the initial identity, we need to show that

$\sin x + \sin (x+120^{\circ}) +\sin(x-120^{\circ})=0.$

The easiest way to do that is to recourse to complex numbers and Euler's formula $e^{i\theta}=\cos\theta +\sin\theta.$ In complex numbers, the three cube roots of unity add up to $0:$

$1 + e^{i\frac{\pi}{3}}+ + e^{-i\frac{\pi}{3}} = 0.$

This multiplied by any factor still remains $0:$

$e^{ix} + e^{i(x+\frac{\pi}{3})}+ + e^{i(x-\frac{\pi}{3})} = 0.$

But, if a complex number is $0$, both its real and imaginary parts also vanish. For the imaginary part, we have exactly what's needed:

$\displaystyle\sin x + \sin (x+\frac{\pi}{3}) +\sin(x-\frac{\pi}{3})=0.$


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