Question
For $\triangle A B C$, prove that:
$
\sin \left(\frac{A+B}{2}\right)=\cos \frac{C}{2}
$
$
\sin \left(\frac{A+B}{2}\right)=\cos \frac{C}{2}
$
✓
Answer
$
\sin \left(\frac{A+B}{2}\right)=\cos \frac{C}{2}
$
We know that for a triangle $\triangle A B C$
$
\begin{aligned}
& <A+<B+<C=180^{\circ} \\
& \frac{<B+<A}{2}=90^{\circ}-\frac{<C}{2} \\
& \sin \left(\frac{A+B}{2}\right)=\sin \left(90^{\circ}-\frac{C}{2}\right) \\
& =\cos \left(\frac{C}{2}\right)
\end{aligned}
$
\sin \left(\frac{A+B}{2}\right)=\cos \frac{C}{2}
$
We know that for a triangle $\triangle A B C$
$
\begin{aligned}
& <A+<B+<C=180^{\circ} \\
& \frac{<B+<A}{2}=90^{\circ}-\frac{<C}{2} \\
& \sin \left(\frac{A+B}{2}\right)=\sin \left(90^{\circ}-\frac{C}{2}\right) \\
& =\cos \left(\frac{C}{2}\right)
\end{aligned}
$
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