Solution:
sinα +sin(120°+α) + sin(240°+α)
= sinα+ 2 sin $\frac{(120° + {\mathrm{\alpha}} + 240°+ {\mathrm{\alpha}})}{2}$ cos$\frac{(120° + {\mathrm{\alpha}} – 240° – {\mathrm{\alpha}})}{2}$
= sinα + 2 sin $\frac{360° + 2{\mathrm{\alpha}}}{2}$ cos$\left( – \frac{120}{2}\right)$
= sinα + 2 sin (180°+α) cos 60°
= sinα – 2 sinα × $\frac{1}{2}$ [∵ sin (180°+α) = – sinα]
= sinα-sinα
= 0
∴ sinα +sin(120°+α) + sin(240°+α) = 0 [Proved]