Question

Find the principal values of each of the following:

(i) ${\sec }^{-1}(-\sqrt{2})$
(ii) ${\sec }^{-1}\left(2\right)$
(iii) ${\sec }^{-1}\left(2\sin \frac{3\mathrm{\pi }}{4}\right)$
(iv) ${\sec }^{-1}\left(2\tan \frac{3\mathrm{\pi }}{4}\right)$

Answer 1

(i) Let ${\sec }^{-1}\left(-\sqrt{2}\right)=y$
Then,
$\sec y=-\sqrt{2}$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]-\left\{\frac{\mathrm{\pi }}{2}\right\}$.
Thus,
$$\begin{gathered} \sec y=-\sqrt{2}=\sec \left(\frac{3\mathrm{\pi }}{4}\right) \\ \Rightarrow y=\frac{3\mathrm{\pi }}{4}\in \left[0,\mathrm{\pi }\right],y\ne \frac{\mathrm{\pi }}{2} \end{gathered}$$
Hence, the principal value of ${\sec }^{-1}\left(-\sqrt{2}\right)\mathrm{is}\frac{3\mathrm{\pi }}{4}$.

(ii) Let ${\sec }^{-1}\left(2\right)=y$
Then,
$\sec y=2$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]-\left\{\frac{\mathrm{\pi }}{2}\right\}$.
Thus,
$$\begin{gathered} \sec y=2=\sec \left(\frac{\mathrm{\pi }}{3}\right) \\ \Rightarrow y=\frac{\mathrm{\pi }}{3}\in \left[0,\mathrm{\pi }\right],y\ne \frac{\mathrm{\pi }}{2} \end{gathered}$$
Hence, the principal value of ${\sec }^{-1}\left(2\right)\mathrm{is}\frac{\mathrm{\pi }}{3}$.

(iii)
Let ${\sec }^{-1}\left(2\sin \frac{3\mathrm{\pi }}{4}\right)=y$
Then,
$\sec y=2\sin \frac{3\mathrm{\pi }}{4}$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]-\left\{\frac{\mathrm{\pi }}{2}\right\}$.
Thus,
$$\begin{gathered} \sec y=2\sin \frac{3\mathrm{\pi }}{4}=2\times \frac{1}{\sqrt{2}}=\sqrt{2}=\sec \left(\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow y=\frac{\mathrm{\pi }}{4}\in \left[0,\mathrm{\pi }\right] \end{gathered}$$
Hence, the principal value of ${\sec }^{-1}\left(2\sin \frac{3\mathrm{\pi }}{4}\right)\mathrm{is}\frac{\mathrm{\pi }}{4}$.
(iv)
Let ${\sec }^{-1}\left(2\tan \frac{3\mathrm{\pi }}{4}\right)=y$
Then,
$\sec y=2\tan \frac{3\mathrm{\pi }}{4}$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]-\left\{\frac{\mathrm{\pi }}{2}\right\}$.
Thus,
$$\begin{gathered} \sec y=2\tan \frac{3\mathrm{\pi }}{4}=2\times \left(-1\right)=-2=\sec \left(\frac{2\mathrm{\pi }}{3}\right) \\ \Rightarrow y=\frac{2\mathrm{\pi }}{3}\in \left[0,\mathrm{\pi }\right] \end{gathered}$$
Hence, the principal value of ${\sec }^{-1}\left(2\tan \frac{3\mathrm{\pi }}{4}\right)\mathrm{is}\frac{2\mathrm{\pi }}{3}$.