Question

If $\cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}=\sin \theta_{1}+\sin \theta_{2}+\sin \theta_{3}=0$, then the value of $\cos \left(\theta_{1}+\theta_{2}\right)+\cos \left(\theta_{2}+\theta_{3}\right)+\cos \left(\theta_{3}+\theta_{1}\right)$ is

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $0$
• D. $1$

Answer 1

Answer: (C)

$z_{1}=\cos \theta_{1}+i \sin \theta_{1}=e^{i \theta_{1}}$
$z_{2}=\cos \theta_{2}+i \sin \theta_{2}=e^{i \theta_{2}}$
$z_{3}=\cos \theta_{3}+i \sin \theta_{3}=e^{i \theta_{3}}$
$z_{1}+z_{2}+z_{3}=0$
$\Rightarrow \overline{z_{1}}+\overline{z_{2}}+\overline{z_{3}}=0$
$\Rightarrow \frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}=0$
$\therefore z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}=0$
$\Rightarrow e^{i\left(\theta_{1}+\theta_{2}\right)}+e^{i\left(\theta_{2}+\theta_{3}\right)}+e^{i\left(\theta_{3}+\theta_{1}\right)}=0$
$\therefore \Sigma \cos \left(\theta_{1}+\theta_{2}\right)=0$