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
equal to

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

Answer 1

Answer: (D)

Let $z_1=\cos A+i\sin A$
$z_2=\cos B+i\sin B$
and $z_3=\cos C+i\sin C$
Then, ${\bar{z}}_1=\cos A-i\sin A,$
${\bar{z}}_2=\cos B-i\sin B$
and ${\bar{z}}_3=\cos C-i\sin C$
Now, if $z=\cos \theta +i\sin \theta$
$\bar{z}=\cos \theta -i\sin \theta$
$=\frac{(\cos \theta -i\sin \theta )(\cos \theta +i\sin \theta )}{(\cos \theta +i\sin \theta )}$
$=\frac{{\cos }^2\theta +{\sin }^2\theta }{\cos \theta +i\sin \theta }=\frac{1}{\cos \theta +i\sin \theta }=\frac{1}{z}$ ,
${\bar{z}}_1+{\bar{z}}_2+{\bar{z}}_3=(\cos A-i\sin A)$
$+(\cos B-i\sin B)+(\cos C-i\sin C)$
$=(\cos A+\cos B+\cos C)$
$-i(\sin A+\sin B+\sin C)=0$
$\Rightarrow z_1{z}_2+z_2{z}_3+z_3{z}_1=0$
$\left[\because \bar{z}=\frac{1}{z}\right]$
$\Rightarrow \Sigma (\cos A+i\sin A)(\cos B+i\sin B)=0$
$\Sigma (\cos A\cos B-\sin A\sin B)$
$+i(\sin A\cos B+\cos A\sin B)=0$

$\Rightarrow \Sigma \cos (A+B)+i\sin (A+B)=0$
$\sum \cos (A+B)=0$
$\Rightarrow \cos (A+B)+\cos (B+C)+\cos (C+A)=0$