Q.If $\cos\alpha + \cos\beta + \cos\gamma = 0 = \sin\alpha + \sin\beta + \sin\gamma$, prove that : $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = \dfrac{3}{2} = \sin^2\alpha + \sin^2\beta + \sin^2\gamma.$

Treat $\cos\theta+i\sin\theta$ as unit vectors; the given conditions say three unit vectors sum to zero, which forces them to be spaced $120^\circ$ apart — and that symmetry makes both cosine-squared and sine-squared sums equal $\tfrac32$.

Let $z_1=\cos\alpha+i\sin\alpha$, $z_2=\cos\beta+i\sin\beta$, $z_3=\cos\gamma+i\sin\gamma$ — each has $|z_k|=1$.

The given conditions combine into one complex equation:

$z_1+z_2+z_3 = (\cos\alpha+\cos\beta+\cos\gamma) + i(\sin\alpha+\sin\beta+\sin\gamma) = 0+i\cdot0 = 0$

Step 1 — pairwise angle differences. From $z_1+z_2=-z_3$, taking modulus-squared of both sides:

$|z_1+z_2|^2 = |z_3|^2=1$

$(z_1+z_2)(\overline{z_1}+\overline{z_2}) = 1 \implies |z_1|^2+|z_2|^2+z_1\overline{z_2}+\overline{z_1}z_2 = 1$

$1+1+2\cos(\alpha-\beta) = 1 \implies \cos(\alpha-\beta) = -\frac12$

By the same argument on any pair, $\cos(\alpha-\beta)=\cos(\beta-\gamma)=\cos(\gamma-\alpha)=-\dfrac12$, i.e. every pairwise angle differs by exactly $120^\circ$ ($\tfrac{2\pi}3$). Three unit vectors can only sum to zero if they are mutually $120^\circ$ apart — so $\alpha,\beta,\gamma$ are, in some order, $\theta,\ \theta+120^\circ,\ \theta+240^\circ$ for some $\theta$.

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