Question: Write the principal value of $\cos ^{-1}\left(\cos 680^{\circ}\right)$
Solution:
$\cos ^{-1}\left(\cos 680^{\circ}\right)=\cos ^{-1}\left[\cos \left(720^{\circ}-680^{\circ}\right)\right]$
$=\cos ^{-1}\left(\cos 40^{\circ}\right)$
$=40^{\circ}$
