Question

Write the principal value (in degrees) of ${\cos }^{-1}\left[\cos \right(680{)}^{\circ }]$

Answer 1

${680}^{\circ }$ is not in principal interval of $\cos$ which is $[0,{180}^{\circ }]$

$\therefore$ by reducing the angle within the principal value,

by using $\cos (n\times {360}^{\circ }-\theta )=\cos \theta$

$\therefore {\cos }^{-1}\left(\cos {680}^{\circ }\right)={\cos }^{-1}\cos ({720}^{\circ }-{40}^{\circ })={\cos }^{-1}\cos (2\times {360}^{\circ }-{40}^{\circ })$

$={\cos }^{-1}\left(\cos {40}^{\circ }\right)={40}^{\circ }$

$\therefore {\cos }^{-1}\left[cos\right({680}^o\left)\right]={40}^o$