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Looking for a strictly trigonometric solution for three-phase systems. Trying to find alternate form for:

$$\sin(x)-\sin(x-120^{\circ})$$

From using WolframAlpha for the expansion:

$$\sin(x)-\sin(x)\cos(120^{\circ})-\cos(x)\sin(120^{\circ})$$

showed an alternate form of:

$$\sqrt{2-2\cos(120^{\circ})}\sin\left(x-\tan^{-1}\left(\frac{\sin(120^{\circ})}{1-\cos(120^{\circ})}\right)\right)$$

This is the form of solution I was looking for although using trigonometric identities I am unable to achieve an answer in this form. Maybe a conversion into another co-ordinate system but honestly have no idea how to proceed. Any help on this would be greatly appreciated.

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\begin{align} \sin(x)-\sin(x-120^{\circ}) &= \sin((x - 60^{\circ})+60^{\circ})-\sin((x - 60^{\circ})-60^{\circ})\\ &= 2\cos(x - 60^{\circ})\sin(60^{\circ})\\ &= \sqrt3 \cos(x - 60^{\circ})\\ &= \sqrt 3 (\cos x \cos 60^{\circ} + \sin x \sin 60^{\circ})\\ &= \dfrac{\sqrt 3}{2} \cos x + \dfrac 32 \sin x \end{align}

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$$\begin{align} \sin(x)-\sin(x-120°) &= \sin(x)-(\sin(x)\cos(120°)-\cos(x)\sin(120°)) \\[2 ex] &= \sin(x)(1-\cos(120°))+\cos(x)\sin(120°) \\[2 ex] &= \sin(x)\cdot (1--\frac 12)+\cos(x)\cdot \frac{\sqrt 3}2 \\[2 ex] &= \sin(x)\cdot\frac 32+\cos(x)\cdot\frac{\sqrt 3}2 \\[2 ex] &= \sqrt 3\left( \sin(x)\cdot\frac{\sqrt 3}2+\cos(x)\cdot\frac 12 \right) \\[2 ex] &= \sqrt 3(\sin(x)\cdot\cos(30°)+\cos(x)\cdot\sin(30°)) \\[2 ex] &= \sqrt 3\sin(x+30°) \end{align}$$

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