​Find the principal values of each of the following:

Solution:

(i) Let $\operatorname{cosec}^{-1}(-\sqrt{2})=y$

Then,

cosec $y=-\sqrt{2}$

We know that the range of the principal value branch is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$.

Thus,

$\operatorname{cosec} y=-\sqrt{2}=\operatorname{cosec}\left(-\frac{\pi}{4}\right)$

$y=-\frac{\pi}{4} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], y \neq 0$

Hence, the principal value of $\operatorname{cosec}^{-1}(-\sqrt{2})$ is $-\frac{\pi}{4}$.

(ii)

Let $\operatorname{cosec}^{-1}(-2)=y$

Then,

cosec $y=-2$

We know that the range of the principal value branch is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$.

Thus,

$\operatorname{cosec} y=-2=\operatorname{cosec}\left(-\frac{\pi}{6}\right)$

$y=-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], y \neq 0$

Hence, the principal value of $\operatorname{cosec}^{-1}(-2)$ is $-\frac{\pi}{6}$.

(iii) Let $\operatorname{cosec}^{-1}\left(\frac{2}{\sqrt{3}}\right)=y$

Then,

$\operatorname{cosec} y=\frac{2}{\sqrt{3}}$

We know that the range of the principal value branch is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$.

Thus,

$\operatorname{cosec} y=\frac{2}{\sqrt{3}}=\operatorname{cosec}\left(\frac{\pi}{3}\right)$

$\Rightarrow y=\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], y \neq 0$

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

(iv)

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

Then,

$\operatorname{cosec} y=2 \cos \frac{2 \pi}{3}$

We know that the range of the principal value branch is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$.

Thus,

$\operatorname{cosec} y=2 \cos \frac{2 \pi}{3}=2 \times \frac{-1}{2}=-1=\operatorname{cosec}\left(-\frac{\pi}{2}\right)$

$\Rightarrow y=-\frac{\pi}{2} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], y \neq 0$

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