Question

Evaluate the following:
(i) ${\sin }^{-1}\left(\sin \frac{5\mathrm{\pi }}{6}\right)$
(ii) ${\cos }^{-1}\left\{\cos \left(-\frac{\mathrm{\pi }}{4}\right)\right\}$
(iii) ${\tan }^{-1}\left\{\tan \frac{3\mathrm{\pi }}{4}\right\}$
(iv) ${\sin }^{-1}(\sin 2)$
(v) $\sin \left\{\frac{\mathrm{\pi }}{3}-{\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right\}$
(vi) $\cos \left\{{\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)+\frac{\mathrm{\pi }}{4}\right\}$
(vii) $\cos \left({\tan }^{-1}\frac{3}{4}\right)$
(viii) ${\cos }^{-1}\left\{\cos \frac{5\mathrm{\pi }}{4}\right\}$
(ix) ${\cos }^{-1}\left\{\cos \left(\frac{4\mathrm{\pi }}{3}\right)\right\}$
(x) ${\tan }^{-1}\left(\tan \frac{2\mathrm{\pi }}{3}\right)$
(xi) ${\cos }^{-1}\left(\cos \frac{13\mathrm{\pi }}{6}\right)$
(xii) ${\tan }^{-1}\left(\tan \frac{7\mathrm{\pi }}{6}\right)$

Answer 1

We know

$$\begin{gathered} \sin \theta =\theta \mathrm{if}-\frac{\mathrm{\pi }}{2}\le \theta \le \frac{\mathrm{\pi }}{2} \\ \cos \theta =\theta \mathrm{if}0\le \theta \le \mathrm{\pi } \\ \tan \mathrm{\theta }=\mathrm{\theta }\mathrm{i}\mathrm{f}-\frac{\mathrm{\pi }}{2}<\mathrm{\theta }<\frac{\mathrm{\pi }}{2} \end{gathered}$$

(i) We have

$$\begin{gathered} {\sin }^{-1}\left(\sin \frac{5\mathrm{\pi }}{6}\right)={\sin }^{-1}\left\{\sin \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{6}\right)\right\} \\ ={\sin }^{-1}\left(\sin \frac{\mathrm{\pi }}{6}\right) \\ =\frac{\mathrm{\pi }}{6} \end{gathered}$$

(ii) We have

$$\begin{gathered} {\cos }^{-1}\left\{\cos \left(-\frac{\mathrm{\pi }}{4}\right)\right\}={\cos }^{-1}\left\{\cos \left(\frac{\mathrm{\pi }}{4}\right)\right\} \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$


(iii) We have

$$\begin{gathered} {\tan }^{-1}\left\{\tan \left(\frac{3\mathrm{\pi }}{4}\right)\right\}={\tan }^{-1}\left\{\tan \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right)\right\} \\ ={\tan }^{-1}\left\{-\tan \left(\frac{\mathrm{\pi }}{4}\right)\right\} \\ ={\tan }^{-1}\left\{\tan \left(-\frac{\mathrm{\pi }}{4}\right)\right\} \\ =-\frac{\mathrm{\pi }}{4} \end{gathered}$$


(iv) We have

$$\begin{gathered} {\sin }^{-1}\left(\sin 2\right)={\sin }^{-1}\left\{\sin \left(\mathrm{\pi }-2\right)\right\}\left[\because 2\notin \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\right] \\ =\mathrm{\pi }-2 \end{gathered}$$


(v) We have

$$\begin{gathered} \sin \left\{\frac{\pi }{3}-{\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right\}=\sin \left\{\frac{\pi }{3}-\mathrm{si}{n}^{-1}\left(\sin \left(-\frac{\pi }{3}\right)\right)\right\}\left[\because -\frac{\pi }{3}\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]\right] \\ =\sin \left\{\frac{\pi }{3}-\left(-\frac{\pi }{3}\right)\right\} \\ =\sin \frac{2\pi }{3} \\ =\frac{\sqrt{3}}{2} \end{gathered}$$
$\therefore \sin \left\{\frac{\pi }{3}-{\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right\}=\frac{\sqrt{3}}{2}$

(vi) We have

$$\begin{gathered} \cos \left\{{\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)+\frac{\mathrm{\pi }}{4}\right\}=\cos \left\{{\cos }^{-1}\left(\cos \left(\frac{5\mathrm{\pi }}{6}\right)+\frac{\mathrm{\pi }}{4}\right)\right\} \\ =\cos \left\{\frac{5\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{4}\right\} \\ =\cos \frac{13\mathrm{\pi }}{12} \\ =\cos \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{12}\right) \\ =-\cos \frac{\mathrm{\pi }}{12} \\ =-\cos \left(\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{6}\right) \\ =-\left\{\cos \frac{\mathrm{\pi }}{4}\cos \frac{\mathrm{\pi }}{6}+\sin \frac{\mathrm{\pi }}{4}\sin \frac{\mathrm{\pi }}{6}\right\} \\ =-\left\{\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}\right\} \\ =-\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right) \end{gathered}$$


(vii) We have

$$\begin{gathered} \cos \left({\tan }^{-1}\frac{3}{4}\right)=\cos \left[\frac{1}{2}{\cos }^{-1}\left(\frac{1-{\left({\displaystyle \frac{3}{4}}\right)}^2}{1+{\left(\frac{3}{4}\right)}^2}\right)\right]\left[\because 2{\tan }^{-1}x={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] \\ =\cos \left[\frac{1}{2}{\cos }^{-1}\left(\frac{7}{25}\right)\right] \end{gathered}$$
Let $$\begin{gathered} y={\cos }^{-1}\left(\frac{7}{25}\right) \\ \Rightarrow \cos y=\frac{7}{25} \end{gathered}$$
Now,

$$\begin{gathered} \cos \left[\frac{1}{2}{\cos }^{-1}\left(\frac{7}{25}\right)\right]=\cos \left[\frac{1}{2}y\right] \\ =\sqrt{\frac{\cos y+1}{2}}\left[\because \cos 2x=2{\cos }^{2}x-1\right] \\ =\sqrt{\frac{{\displaystyle \frac{7}{25}}+1}{2}} \\ =\sqrt{\frac{32}{50}} \\ =\frac{4}{5} \end{gathered}$$
$\therefore \cos \left[{\tan }^{-1}\left(\frac{3}{4}\right)\right]=\frac{4}{5}$

(viii) We have

$$\begin{gathered} {\cos }^{-1}\left\{\cos \left(\frac{5\mathrm{\pi }}{4}\right)\right\}={\cos }^{-1}\left\{\cos \left(2\mathrm{\pi }-\frac{3\mathrm{\pi }}{4}\right)\right\} \\ ={\cos }^{-1}\left\{\cos \left(\frac{3\mathrm{\pi }}{4}\right)\right\} \\ =\frac{3\mathrm{\pi }}{4} \end{gathered}$$

(ix) We have

$$\begin{gathered} {\cos }^{-1}\left\{\cos \left(\frac{4\mathrm{\pi }}{3}\right)\right\}={\cos }^{-1}\left\{\cos \left(2\mathrm{\pi }-\frac{2\mathrm{\pi }}{3}\right)\right\} \\ ={\cos }^{-1}\left\{\cos \left(\frac{2\mathrm{\pi }}{3}\right)\right\} \\ =\frac{2\mathrm{\pi }}{3} \end{gathered}$$


(x) We have

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

(xi) We have

$$\begin{gathered} {\cos }^{-1}\left\{\cos \left(\frac{13\mathrm{\pi }}{6}\right)\right\}={\cos }^{-1}\left\{\cos \left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right\} \\ ={\cos }^{-1}\left\{\cos \left(\frac{\mathrm{\pi }}{6}\right)\right\} \\ =\frac{\mathrm{\pi }}{6} \end{gathered}$$


(xii) We have

$$\begin{gathered} {\tan }^{-1}\left\{\tan \left(\frac{7\mathrm{\pi }}{6}\right)\right\}={\tan }^{-1}\left\{\tan \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right\} \\ ={\tan }^{-1}\left\{\tan \left(\frac{\mathrm{\pi }}{6}\right)\right\} \\ =\frac{\mathrm{\pi }}{6} \end{gathered}$$