If $\sin A+\sin B=\frac{1}{2}$ and $\cos A+\cos B=1$, then $\sin \left(\frac{A-B}{2}\right)$ equals

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Aicharya · Accepted Answer

Given $\sin A+\sin B=\frac{1}{2}$ and $\cos A+\cos B=1$, on square and add the given relations, we get $ \begin{array}{ll} 2+2(\cos A \cos B+\sin A \sin B)=\frac{1}{4}+1 \ \Rightarrow & 2 \cos (A-B)=-\frac{3}{4} \ \Rightarrow & \cos (A-B)=-\frac{3}{8} \end{array} $ $ \begin{array}{lc} \Rightarrow & 1-2 \sin ^2\left(\frac{A-B}{2}ight)=-\frac{3}{8} \Rightarrow 2 \sin ^2\left(\frac{A-B}{2}ight)=\frac{11}{8} \ \Rightarrow & \sin \frac{A-B}{2}= \pm \frac{\sqrt{11}}{4} \end{array} $

If $\sin A+\sin B=\frac{1}{2}$ and $\cos A+\cos B=1$, then $\sin \left(\frac{A-B}{2}\right)$ equals

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