If sin A+sin B+sin C=0 and cos A+cos B+cos C= | vedclass

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(D) Let $z_1 = \cos A + i \sin A$,$z_2 = \cos B + i \sin B$,and $z_3 = \cos C + i \sin C$. Given $\cos A + \cos B + \cos C = 0$ and $\sin A + \sin B +

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(D) Let $z_1 = \cos A + i \sin A$,$z_2 = \cos B + i \sin B$,and $z_3 = \cos C + i \sin C$. Given $\cos A + \cos B + \cos C = 0$ and $\sin A + \sin B + \sin C = 0$,we have $z_1 + z_2 + z_3 = 0$. Taking the conjugate,$\bar{z}_1 + \bar{z}_2 + \bar{z}_3 = 0$. Since $\bar{z} = \frac{1}{z}$ for $|z|=1$,we have $\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} = 0$. This implies $\frac{z_2 z_3 + z_3 z_1 + z_1 z_2}{z_1 z_2 z_3} = 0$,so $z_1 z_2 + z_2 z_3 + z_3 z_1 = 0$. Substituting the polar forms: $\sum (\cos A + i \sin A)(\cos B + i \sin B) = 0$ $\sum (\cos A \cos B - \sin A \sin B) + i \sum (\sin A \cos B + \cos A \sin B) = 0$ $\sum \cos (A+B) + i \sum \sin (A+B) = 0$. Equating the real parts,we get $\cos (A+B) + \cos (B+C) + \cos (C+A) = 0$.

If $\sin A+\sin B+\sin C=0$ and $\cos A+\cos B+\cos C=0$,then $\cos (A+B)+\cos (B+C)+\cos (C+A)$ is equal to

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