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$
$\Rightarrow \cos (A+B)+\cos (B+C)+\cos (C+A)=0$