Question

In a \( \triangle ABC \), prove that :

\[ \cos\frac{A+B}{2} = \sin\frac{C}{2} \]

Solution

In a triangle,

\[ A+B+C=\pi \]

\[ A+B=\pi-C \]

Dividing by \(2\),

\[ \frac{A+B}{2} = \frac{\pi-C}{2} = \frac{\pi}{2}-\frac{C}{2} \]

Therefore,

\[ \begin{aligned} \cos\frac{A+B}{2} &= \cos\left(\frac{\pi}{2}-\frac{C}{2}\right) \\[8pt] &= \sin\frac{C}{2} \end{aligned} \]

Hence Proved.