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


From using WolframAlpha for the expansion:


showed an alternate form of:


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.


2 Answers 2


\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}


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.