$\underline{\textbf{Solution 1}}$ One way to prove these two identity is simply to expand: \begin{align} &\sin x +\sin ( x+120^\circ) + \sin (x-120^\circ) \\ &=\sin x + (\sin x \cos 120^\circ + \cos x \sin 120^\circ)+(\sin x \cos 120^\circ - \cos x \sin 120^\circ)\\ &= \sin x + 2\sin x\cos 120^\circ\\ &=\sin x + 2\sin x \cdot (-\frac{1}{2})\\ &=0 \end{align} \begin{align} &\cos x +\cos ( x+120^\circ) + \cos (x - 120^\circ) \\ &=\cos x + (\cos x \cos 120^\circ + \sin x \sin 120^\circ)+(\cos x \cos 120^\circ - \sin x \sin 120^\circ)\\ &= \cos x + 2\cos x\cos 120^\circ\\ &=\cos x + 2\cos x \cdot (-\frac{1}{2})\\ &=0 \end{align} $\underline{\textbf{Solution 2}}$ Let $z_1=\cos x +i\sin x$, $z_2 = \cos(x+120^\circ)+i\sin(x+120^\circ)$, and $z_3=\cos (x-120^\circ) + i\sin(x-120^\circ)$. Then $z_1$, $z_2$, and $z_3$ are three unit vectors which can form an equilateral triangle. Hence $z_1 + z_2 + z_3=0$. This means both of its real part and imaginary part equal $0$.