If A + B =
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

$\Rightarrow 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)=1 \quad\left[\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$

$\Rightarrow 2 \cos \left(\frac{\pi}{6}\right) \cos \left(\frac{A-B}{2}\right)=1$   $\left[\because A+B=\frac{\pi}{3}\right]$

 

$\Rightarrow 2 \times \frac{\sqrt{3}}{2} \times \cos \left(\frac{A-B}{2}\right)=1$

$\Rightarrow \cos \left(\frac{A-B}{2}\right)=\frac{1}{\sqrt{3}}$

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