Question

If $\sin A+\sin B+\sin C=0$ and $\cos A+\cos B+\cos C=0$, then the value of $\sin \left(\frac{A-B}{2}\right)$ is
$(\text{where}A,B,C\in [0,2\pi ])$

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is D $\frac{\sqrt{3}}{2}$

$\sin A+\sin B+\sin C=0\Rightarrow \sin A+\sin B=-\sin C\cdots \left(1\right)$
$\cos A+\cos B+\cos C=0\Rightarrow \cos A+\cos B=-\cos C\cdots \left(2\right)$
Squaring and adding both the equation $\left(1\right)$ and $\left(2\right)$,
$(\sin A+\sin B{)}^2+(\cos A+\cos B{)}^2={\sin }^{2}C+{\cos }^{2}C\Rightarrow (\sin A+\sin B{)}^2+(\cos A+\cos B{)}^2=1\Rightarrow {\sin }^{2}A+{\sin }^{2}B+2\sin A\sin B+{\cos }^{2}A+{\cos }^{2}B+2\cos A\cos B=1\Rightarrow 1+1+2\left(\cos \right(A-B\left)\right)=1\Rightarrow \cos (A-B)=-\frac{1}{2}\Rightarrow A-B={120}^{\circ }\text{or}{240}^{\circ }\Rightarrow \frac{A-B}{2}={60}^{\circ }\text{or}{120}^{\circ }\therefore \sin \left(\frac{A-B}{2}\right)=\frac{\sqrt{3}}{2}$