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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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1242 Upvotes
Verified Answer
The correct answer is:
$\pm \frac{\sqrt{11}}{4}$
Given $\sin A+\sin B=\frac{1}{2}$ and $\cos A+\cos B=1$, on square and add the given relations, we get
$$
\begin{array}{ll}
2+2(\cos A \cos B+\sin A \sin B)=\frac{1}{4}+1 \\
\Rightarrow & 2 \cos (A-B)=-\frac{3}{4} \\
\Rightarrow & \cos (A-B)=-\frac{3}{8}
\end{array}
$$
$$
\begin{array}{lc}
\Rightarrow & 1-2 \sin ^2\left(\frac{A-B}{2}\right)=-\frac{3}{8} \Rightarrow 2 \sin ^2\left(\frac{A-B}{2}\right)=\frac{11}{8} \\
\Rightarrow & \sin \frac{A-B}{2}= \pm \frac{\sqrt{11}}{4}
\end{array}
$$
$$
\begin{array}{ll}
2+2(\cos A \cos B+\sin A \sin B)=\frac{1}{4}+1 \\
\Rightarrow & 2 \cos (A-B)=-\frac{3}{4} \\
\Rightarrow & \cos (A-B)=-\frac{3}{8}
\end{array}
$$
$$
\begin{array}{lc}
\Rightarrow & 1-2 \sin ^2\left(\frac{A-B}{2}\right)=-\frac{3}{8} \Rightarrow 2 \sin ^2\left(\frac{A-B}{2}\right)=\frac{11}{8} \\
\Rightarrow & \sin \frac{A-B}{2}= \pm \frac{\sqrt{11}}{4}
\end{array}
$$
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