Question

If A + B = $\frac{\mathrm{\pi }}{3}$ and cos A + cos B = 1, then find the value of cos $\frac{A-B}{2}$.

Answer 1

Given:
A + B = $\frac{\mathrm{\pi }}{3}$
and cos A + cos B = 1

$$\begin{gathered} \Rightarrow 2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)=1\left[\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right] \\ \Rightarrow 2\cos \left(\frac{\mathrm{\pi }}{6}\right)\cos \left(\frac{A-B}{2}\right)=1\left[\because A+B=\frac{\mathrm{\pi }}{3}\right] \\ \Rightarrow 2\times \frac{\sqrt{3}}{2}\times \cos \left(\frac{A-B}{2}\right)=1 \\ \Rightarrow \cos \left(\frac{A-B}{2}\right)=\frac{1}{\sqrt{3}} \end{gathered}$$