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

Here, we have the pattern of data in $\cos x$ and $\sin x$, so let's use the method of polar form of complex number. Polar form of complex number, $z=\cos \theta+i \sin \theta$ then $\bar{z}=\cos \theta-i \sin \theta$ and $z \bar{z}=1$ $\Rightarrow \quad \bar{z}=\frac{1}{z}$ Let $z_1=\cos A+i \sin A$ $z_2=\cos B+i \sin B$ and $z_3=\cos C+i \sin C$ Then, $\quad \bar{z}_1=\cos A-i \sin A$ $\begin{gathered} \bar{z}_2=\cos B-i \sin B \\ z_3=\cos C-\sin C \end{gathered}$ Now, to find $\cos (A+B)+\cos (B+C)+\cos (C+A)$ $\begin{aligned} & \bar{z}_1+\bar{z}_2+\bar{z}_3=(\cos A-i \sin A) \\ & +(\cos B-i \sin B)+(\cos C-i \sin C) \\ & =(\cos A+\cos B+\cos C)-i(\sin A+\sin B+\sin C)=0 \\ & \frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=0 \quad\left[\because \quad \bar{z}=\frac{1}{z}\right] \\ & z_1 z_2+z_2 z_3+z_3 z_1=0 \\ & \sum(\cos A+i \sin A)(\cos B+i \sin B)=0 \\ & \sum(\cos A \cos B-\sin A \sin B) \\ & +i(\sin A \cos B+\cos A \sin B)=0 \\ & \sum \cos (A+B)+i \sin (A+B)=0 \\ & \sum \cos (A+B)=0 \end{aligned}$ Comparing the real part, we get $\cos (A+B)+\cos (B+C)+\cos (C+A)=0$