Question

For the principal values, evaluate each of the following:

(i) ${\cos }^{-1}\frac{1}{2}+2{\sin }^{-1}\left(\frac{1}{2}\right)$
(ii)
(iii) ${\sin }^{-1}\left(-\frac{1}{2}\right)+2{\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
(iv) ${\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)+{\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Answer 1

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

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

(ii)

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

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