Question

Find the principal values of each of the following:

(i) ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
(ii) ${\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)$

(iii) ${\tan }^{-1}\left(\cos \frac{\mathrm{\pi }}{2}\right)$

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

Answer 1

(i) Let ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=y$
Then,
$\tan y=\frac{1}{\sqrt{3}}$
We know that the range of the principal value branch is $\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$.
Thus,
$$\begin{gathered} \tan y=\frac{1}{\sqrt{3}}=\tan \left(\frac{\mathrm{\pi }}{6}\right) \\ \Rightarrow y=\frac{\mathrm{\pi }}{6}\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right) \end{gathered}$$
Hence, the principal value of ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\mathrm{is}\frac{\mathrm{\pi }}{6}$.

(ii) We have ${\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\left[\because {\tan }^{-1}\left(-x\right)=-{\tan }^{-1}x\right]$
Let ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=y$
Then,
$\tan y=\frac{1}{\sqrt{3}}$
We know that the range of the principal value branch is $\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$.
Thus,
$$\begin{gathered} \tan y=\frac{1}{\sqrt{3}}=\tan \left(\frac{\mathrm{\pi }}{6}\right) \\ \Rightarrow y=\frac{\mathrm{\pi }}{6} \\ \therefore {\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ =-y \\ =-\frac{\mathrm{\pi }}{6}\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right) \end{gathered}$$
Hence, the principal value of ${\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)\mathrm{is}-\frac{\mathrm{\pi }}{6}$.

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

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