Evaluate each of the following:

Solution:

We know

\cos ^{-1}(\cos \theta)=\theta \text { if } 0 \leq \theta \leq \pi

(i)  We have

$\cos ^{-1}\left\{\cos \left(-\frac{\pi}{4}\right)\right\}=\cos ^{-1}\left\{\cos \left(\frac{\pi}{4}\right)\right\}$

$=\frac{\pi}{4}$

(ii) We have

$\cos ^{-1}\left\{\cos \left(\frac{5 \pi}{4}\right)\right\}=\cos ^{-1}\left\{\cos \left(2 \pi-\frac{3 \pi}{4}\right)\right\}$

$=\cos ^{-1}\left\{\cos \left(\frac{3 \pi}{4}\right)\right\}$

$=\frac{3 \pi}{4}$

(iii) We have

$\cos ^{-1}\left\{\cos \left(\frac{4 \pi}{3}\right)\right\}=\cos ^{-1}\left\{\cos \left(2 \pi-\frac{2 \pi}{3}\right)\right\}$

$=\cos ^{-1}\left\{\cos \left(\frac{2 \pi}{3}\right)\right\}$

$=\frac{2 \pi}{3}$

(iv) We have

$\cos ^{-1}\left\{\cos \left(\frac{13 \pi}{6}\right)\right\}=\cos ^{-1}\left\{\cos \left(2 \pi+\frac{\pi}{6}\right)\right\}$

$=\cos ^{-1}\left\{\cos \left(\frac{\pi}{6}\right)\right\}$

$=\frac{\pi}{6}$

(v) We have

$\cos ^{-1}(\cos 3)=3$

(vi)We have

$\cos ^{-1}(\cos 4)=\cos ^{-1}\{\cos (2 \pi-4)\}$

$=2 \pi-4$

(vii) We have

$\cos ^{-1}(\cos 5)=\cos ^{-1}\{\cos (2 \pi-5)\}$

$=2 \pi-5$

(viii) We have

$\cos ^{-1}(\cos 12)=\cos ^{-1}\{\cos (4 \pi-12)\}$

$=4 \pi-12$