Question

Evaluate each of the following:
(i) ${\cos }^{-1}\frac{1}{2}+2{\sin }^{-1}\left(\frac{1}{2}\right)$
(ii) ${\tan }^{-1}\left\{2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right\}$
(iii) ${\tan }^{-1}1+{\cos }^{-1}\left(-\frac{1}{2}\right)+{\sin }^{-1}\left(-\frac{1}{2}\right)$
(iv) ${\tan }^{-1}\sqrt{3}-{\sec }^{-1}(-2)+{\mathrm{cosec}}^{-1}\frac{2}{\sqrt{3}}$

Answer 1

(i)
$$\begin{gathered} {\cos }^{-1}\left(\cos x\right)=x \\ {\sin }^{-1}\left(\sin x\right)=x \end{gathered}$$

$$\begin{gathered} {\cos }^{-1}\left(\frac{1}{2}\right)+2{\sin }^{-1}\left(\frac{1}{2}\right) \\ ={\cos }^{-1}\left(\cos \frac{\mathrm{\pi }}{3}\right)+2{\sin }^{-1}\left(\sin \frac{\mathrm{\pi }}{6}\right) \\ =\frac{\mathrm{\pi }}{3}+2\left(\frac{\mathrm{\pi }}{6}\right) \\ =\frac{2\mathrm{\pi }}{3} \end{gathered}$$

(ii) Let ${\sin }^{-1}\frac{1}{2}=y$
Then,
$\sin y=\frac{1}{2}$
We know that the range of the principal value branch is $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$.
Thus,
$$\begin{gathered} \sin y=\frac{1}{2}=\sin \left(\frac{\mathrm{\pi }}{6}\right) \\ \Rightarrow y=\frac{\mathrm{\pi }}{6}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right] \end{gathered}$$
So,
$$\begin{gathered} {\tan }^{-1}\left\{2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right\} \\ ={\tan }^{-1}\left\{2\cos \left(2\frac{\mathrm{\pi }}{6}\right)\right\} \\ ={\tan }^{-1}\left\{2\cos \left(\frac{\mathrm{\pi }}{3}\right)\right\} \\ ={\tan }^{-1}\left\{2\left(\frac{1}{2}\right)\right\} \\ ={\tan }^{-1}1 \\ =\frac{\mathrm{\pi }}{4}\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right) \end{gathered}$$
$\therefore {\tan }^{-1}\left\{2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right\}=\frac{\pi }{3}$

(iii) Let ${\sin }^{-1}\left(-\frac{1}{2}\right)=y$
Then,
$\sin y=-\frac{1}{2}$
We know that the range of the principal value branch is $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$.
Thus,
$$\begin{gathered} \sin y=-\frac{1}{2}=\sin \left(-\frac{\mathrm{\pi }}{6}\right) \\ \Rightarrow y=-\frac{\mathrm{\pi }}{6}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right] \end{gathered}$$
Now,
Let ${\cos }^{-1}\left(-\frac{1}{2}\right)=z$
Then,
$\cos z=-\frac{1}{2}$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]$.
Thus,
$$\begin{gathered} \cos z=-\frac{1}{2}=\cos \left(\frac{2\mathrm{\pi }}{3}\right) \\ \Rightarrow z=\frac{2\mathrm{\pi }}{3}\in \left[0,\mathrm{\pi }\right] \end{gathered}$$
So,
${\tan }^{-1}1+{\cos }^{-1}\left(-\frac{1}{2}\right)+{\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\mathrm{\pi }}{4}+\frac{2\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6}=\frac{3\mathrm{\pi }}{4}$
$\therefore {\tan }^{-1}1+{\cos }^{-1}\left(-\frac{1}{2}\right)+{\sin }^{-1}\left(\frac{1}{2}\right)=\frac{3\mathrm{\pi }}{4}$

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

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