Question

$a=\cos \theta +i\sin \theta ,b=\cos \varphi +i\sin \varphi ,c=\cos \psi +i\sin \psi$ and $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=2$, then $\sin \left(\theta -\varphi \right)+\sin \left(\varphi -\psi \right)+\sin \left(\psi -\theta \right)$ equals

• A. $3$
• B. $-\frac{3}{2}$
• C. $0$
• D. $\frac{3}{2}$

Answer 1

The correct option is C $0$

Using Demoivre's theorem,

$a=\cos \theta +i\sin \theta =e^{i\theta }$

$b=\cos \varphi +i\sin \varphi =e^{i\varphi }$

$c=\cos \psi +i\sin \psi =e^{i\psi }$

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=2+0.i$

$\frac{e^{i\theta }}{e^{i\varphi }}+\frac{e^{i\varphi }}{e^{i\psi }}+\frac{e^{i\psi }}{e^{i\theta }}=2+0.i$

$e^{i(\theta -\varphi )}+e^{i(\varphi -\psi )}+e^{i(\psi -\theta )}=2+0.i$

$\cos (\theta -\varphi )+\cos (\varphi -\psi )+\cos (\psi -\theta )+i\sin (\theta -\varphi )+i\sin (\varphi -psi)+i\sin (\psi -\theta )=2+0.i$

$[\cos (\theta -\varphi )+\cos (\varphi -\psi )+\cos (\psi -\theta )]+i[\sin (\theta -\varphi )+\sin (\varphi -\psi )+\sin (\psi -\theta )]=2+0.i$

Comparing the real and imaginary parts, we get

$\sin (\theta -\varphi )+\sin (\varphi -\psi )+\sin (\psi -\theta )=0$