Question

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

• A. $\cos (A+B+C)$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is D $0$

Let $x=cosA+isinA,y=cosB+isinB,z=cosC+isinC$

$\frac{1}{x}=cosA-isinA=e^{-iA},\frac{1}{y}=cosB-isinB=e^{-iB},\frac{1}{z}=cosC-isinC=e^{-iC}$

$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

$\frac{xy+yz+xz}{xyz}=0$

$\Rightarrow xy+yz+xz=0$

i.e. $e^{i(A+B)}+e^{i(B+C)}+e^{i(C+A)}=0$

Equating the real and imaginary parts, we get $cos(A+B)+cos(B+C)+cos(C+A)=0$