If cos alpha + cos beta + cos gamma = 0 and s | vedclass

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(C) Given: $\cos \alpha + \cos \beta + \cos \gamma = 0$ $(i)$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$ $(ii)$.Let $a = e^{i\alpha}$,$b = e^{i\

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(C) Given: $\cos \alpha + \cos \beta + \cos \gamma = 0$ $(i)$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$ $(ii)$.Let $a = e^{i\alpha}$,$b = e^{i\beta}$,and $c = e^{i\gamma}$.Then $a + b + c = (\cos \alpha + \cos \beta + \cos \gamma) + i(\sin \alpha + \sin \beta + \sin \gamma) = 0 + i(0) = 0$.Since $a, b, c$ are complex numbers on the unit circle,$|a| = |b| = |c| = 1$,so $\bar{a} = 1/a$,$\bar{b} = 1/b$,and $\bar{c} = 1/c$.Taking the conjugate of $a + b + c = 0$,we get $\bar{a} + \bar{b} + \bar{c} = 0$,which implies $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$.This simplifies to $\frac{ab + bc + ca}{abc} = 0$,so $ab + bc + ca = 0$.Now,consider $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = 0$.Since $ab + bc + ca = 0$,we have $a^2 + b^2 + c^2 = 0$.Substituting $a^2 = e^{i2\alpha} = \cos 2\alpha + i\sin 2\alpha$,etc.,we get:$(\cos 2\alpha + \cos 2\beta + \cos 2\gamma) + i(\sin 2\alpha + \sin 2\beta + \sin 2\gamma) = 0$.Equating the real parts,we get $\cos 2\alpha + \cos 2\beta + \cos 2\gamma = 0$.

If $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$,then $\cos 2\alpha + \cos 2\beta + \cos 2\gamma$ equals:

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