Question

What is the value of cos⁻¹$\left(\cos \frac{2\mathrm{\pi }}{3}\right)+{\sin }^{-1}\left(\sin \frac{2\mathrm{\pi }}{3}\right)$?

Answer 1

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