Question
Find the principal value of:
\[ \cos^{-1}(\cos 680^\circ) \]
Solution
First, reduce the angle:
\[ 680^\circ = 360^\circ + 320^\circ \Rightarrow \cos 680^\circ = \cos 320^\circ \]
Now,
\[ 320^\circ = 360^\circ – 40^\circ \Rightarrow \cos 320^\circ = \cos 40^\circ \]
So,
\[ \cos^{-1}(\cos 680^\circ) = \cos^{-1}(\cos 40^\circ) \]
The principal value range of \( \cos^{-1}x \) is:
\[ [0^\circ, 180^\circ] \]
Since \( 40^\circ \) lies in this range,
\[ \cos^{-1}(\cos 40^\circ) = 40^\circ \]
Final Answer:
\[ \boxed{40^\circ} \]
Key Concept
Reduce the angle first and then check if it lies in the principal value range.