Question

For the principal values, evaluate the following:
(i) ${\sin }^{-1}\frac{1}{2}-2{\sin }^{-1}\frac{1}{\sqrt{2}}$
(ii) ${\sin }^{-1}\left(-\frac{1}{2}\right)+2{\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
(iii) ${\tan }^{-1}(-1)+{\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)$
(iv) ${\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)+{\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Answer 1

(i)
$$\begin{gathered} {\sin }^{-1}\frac{1}{2}-2{\sin }^{-1}\frac{1}{\sqrt{2}} \\ ={\sin }^{-1}\left(\sin \frac{\mathrm{\pi }}{6}\right)-2{\sin }^{-1}\left(\sin \frac{\mathrm{\pi }}{4}\right)\left[\because \mathrm{Range}\mathrm{of}\mathrm{sine}\mathrm{is}\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\mathrm{and}\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{4}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\right] \\ =\frac{\mathrm{\pi }}{6}-2\left(\frac{\mathrm{\pi }}{4}\right) \\ =\frac{\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{2} \\ =-\frac{\mathrm{\pi }}{3} \end{gathered}$$
$\therefore {\sin }^{-1}\frac{1}{2}-2{\sin }^{-1}\frac{1}{\sqrt{2}}=-\frac{\mathrm{\pi }}{3}$
(ii)

$$\begin{gathered} {\sin }^{-1}\left(-\frac{1}{2}\right)+2{\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right) \\ ={\sin }^{-1}\left\{\sin \left(-\frac{\mathrm{\pi }}{6}\right)\right\}+2{\cos }^{-1}\left(\cos \frac{5\mathrm{\pi }}{6}\right)\left[\because \mathrm{Range}\mathrm{of}\mathrm{sine}\mathrm{is}\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right];-\frac{\mathrm{\pi }}{6}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right] \\ \mathrm{and}\mathrm{range}\mathrm{of}\mathrm{cosine}\mathrm{is}\left[0,\mathrm{\pi }\right];\frac{5\mathrm{\pi }}{6}\in \left[0,\mathrm{\pi }\right]\right] \\ =-\frac{\mathrm{\pi }}{6}+2\left(\frac{5\mathrm{\pi }}{6}\right) \\ =-\frac{\mathrm{\pi }}{6}+\frac{5\mathrm{\pi }}{3} \\ =\frac{9\mathrm{\pi }}{6} \\ =\frac{3\mathrm{\pi }}{2} \end{gathered}$$
$\therefore {\sin }^{-1}\left(-\frac{1}{2}\right)+2{\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\frac{3\mathrm{\pi }}{2}$

(iii)
$$\begin{gathered} {\tan }^{-1}\left(-1\right)+{\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right) \\ ={\tan }^{-1}\left\{\tan \left(-\frac{\mathrm{\pi }}{4}\right)\right\}+{\cos }^{-1}\left(\cos \frac{3\mathrm{\pi }}{4}\right)\left[\because \mathrm{Range}\mathrm{of}\tan \mathrm{is}\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right);-\frac{\mathrm{\pi }}{4}\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right) \\ \mathrm{and}\mathrm{range}\mathrm{of}\mathrm{cosine}\mathrm{is}\left[0,\mathrm{\pi }\right];\frac{3\mathrm{\pi }}{4}\in \left[0,\mathrm{\pi }\right]\right] \\ =-\frac{\mathrm{\pi }}{4}+\frac{3\mathrm{\pi }}{4} \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$
$\therefore {\tan }^{-1}\left(-1\right)+{\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{\mathrm{\pi }}{2}$

(iv)

$$\begin{gathered} {\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)+{\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right) \\ ={\sin }^{-1}\left\{\sin \left(-\frac{\mathrm{\pi }}{3}\right)\right\}+{\cos }^{-1}\left(\cos \frac{\mathrm{\pi }}{6}\right) \\ =-\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6}\left[\because \mathrm{Range}\mathrm{of}\mathrm{sine}\mathrm{is}\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right];-\frac{\mathrm{\pi }}{3}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right] \\ \mathrm{and}\mathrm{range}\mathrm{of}\mathrm{cosine}\mathrm{is}\left[0,\mathrm{\pi }\right];\frac{\mathrm{\pi }}{6}\in \left[0,\mathrm{\pi }\right]\right] \\ =-\frac{\mathrm{\pi }}{6} \end{gathered}$$
$\therefore {\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)+{\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)=-\frac{\mathrm{\pi }}{6}$