​Find the principal values of each of the following:

Solution:

(i) Let $\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=y$

Then,

$\cos y=-\frac{\sqrt{3}}{2}$

We know that the range of the principal value branch is $[0, \pi]$.

Thus,

$\cos y=-\frac{\sqrt{3}}{2}=\cos \left(\frac{5 \pi}{6}\right)$

$\Rightarrow y=\frac{5 \pi}{6} \in[0, \pi]$

Hence, the principal value of $\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ is $\frac{5 \pi}{6}$.

(ii) Let $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=y$

Then,

$\cos y=-\frac{1}{\sqrt{2}}$

We know that the range of the principal value branch is [0, π]0, π.

Thus,

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

$\Rightarrow y=\frac{3 \pi}{4} \in[0, \pi]$

Hence, the principal value of $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ is $\frac{3 \pi}{4}$.

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

Then,

$\cos y=\sin \frac{4 \pi}{3}$

We know that the range of the principal value branch is $[0, \pi]$.

Thus,

$\cos y=\sin \frac{4 \pi}{3}=-\frac{\sqrt{3}}{2}=\cos \left(\frac{5 \pi}{6}\right)$

$\Rightarrow y=\frac{5 \pi}{6} \in[0, \pi]$

Hence, the principal value of $\cos ^{-1}\left(\sin \frac{4 \pi}{3}\right)$ is $\frac{5 \pi}{6}$.

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

Then,

$\cos y=\tan \frac{3 \pi}{4}$

We know that the range of the principal value branch is $[0, \pi]$.

Thus,

$\cos y=\tan \frac{3 \pi}{4}=-1=\cos (\pi)$

$\Rightarrow y=\pi \in[0, \pi]$

Hence, the principal value of $\cos ^{-1}\left(\tan \frac{3 \pi}{4}\right)$ is $\pi$.