If \( A+B=\frac{\pi}{3} \) and \( \cos A+\cos B=1 \), then find the value of \( \cos\frac{A-B}{2} \)
Solution:
Using identity,
\[
\cos A+\cos B
=
2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\]
\[
1
=
2\cos\frac{\pi/3}{2}\cos\frac{A-B}{2}
\]
\[
=
2\cos\frac{\pi}{6}\cos\frac{A-B}{2}
\]
Since,
\[
\cos\frac{\pi}{6}=\frac{\sqrt3}{2}
\]
\[
1
=
2\left(\frac{\sqrt3}{2}\right)\cos\frac{A-B}{2}
\]
\[
1
=
\sqrt3\cos\frac{A-B}{2}
\]
\[
\cos\frac{A-B}{2}
=
\frac1{\sqrt3}
\]
\[
\boxed{\frac1{\sqrt3}}
\]