Question:
If $A+B=\frac{\pi}{3}$ and $\cos A+\cos B=1$, then find the value of $\cos \frac{A-B}{2}$.
Solution:
Given:
$A+B=\frac{\pi}{3}$
and cos A + cos B = 1
$\Rightarrow 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)=1 \quad\left[\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$\Rightarrow 2 \cos \left(\frac{\pi}{6}\right) \cos \left(\frac{A-B}{2}\right)=1$ $\left[\because A+B=\frac{\pi}{3}\right]$
$\Rightarrow 2 \times \frac{\sqrt{3}}{2} \times \cos \left(\frac{A-B}{2}\right)=1$
$\Rightarrow \cos \left(\frac{A-B}{2}\right)=\frac{1}{\sqrt{3}}$