Question

The principal value of ${\cos }^{-1}\left(\frac{1}{\sqrt{2}}\left[\cos \left(\frac{9\mathrm{\pi }}{10}\right)-\sin \left(\frac{9\mathrm{\pi }}{10}\right)\right]\right)$ is equal to

Answer 1

Finding the principal value of ${\cos }^{-1}\left(\frac{1}{\sqrt{2}}\left[\cos \left(\frac{9\mathrm{\pi }}{10}\right)-\sin \left(\frac{9\mathrm{\pi }}{10}\right)\right]\right)$:

$\begin{array}{rcl}{\cos }^{-1}\left(\frac{1}{\sqrt{2}}\left[\cos \left(\frac{9\mathrm{\pi }}{10}\right)-\sin \left(\frac{9\mathrm{\pi }}{10}\right)\right]\right)& =& {\cos }^{-1}\left(\frac{1}{\sqrt{2}}\cos \left(\frac{9\mathrm{\pi }}{10}\right)-\frac{1}{\sqrt{2}}\sin \left(\frac{9\mathrm{\pi }}{10}\right)\right)\\ & =& {\cos }^{-1}\left(\cos \left(\frac{\mathrm{\pi }}{4}\right)\cos \left(\frac{9\mathrm{\pi }}{10}\right)-\sin \left(\frac{\mathrm{\pi }}{4}\right)\sin \left(\frac{9\mathrm{\pi }}{10}\right)\right)\mathbf{(}\mathbf{}\mathbf{}\mathbf{\because }\mathbf{cos}\left(\frac{\pi }{4}\right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{sin}\left(\frac{\pi }{4}\right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{)}\\ & =& {\cos }^{-1}\left(\cos \left(\frac{\mathrm{\pi }}{4}+\frac{9\mathrm{\pi }}{10}\right)\right)\mathbf{(}\mathbf{\because }\mathbf{cos}\left(A+B\right)\mathbf{=}\mathbf{cos}\left(A\right)\mathbf{cos}\left(B\right)\mathbf{-}\mathbf{sin}\left(A\right)\mathbf{sin}\left(B\right)\mathbf{)}\\ & =& {\cos }^{-1}\left(\cos \left(\frac{23\mathrm{\pi }}{20}\right)\right)\\ & =& {\cos }^{-1}\left(\cos \left(\frac{20\mathrm{\pi }+3\mathrm{\pi }}{20}\right)\right)\\ & =& {\cos }^{-1}\left(\cos \left(\frac{20\mathrm{\pi }}{20}+\frac{3\mathrm{\pi }}{20}\right)\right)\\ & =& {\cos }^{-1}\left(\cos \left(\mathrm{\pi }+\frac{3\mathrm{\pi }}{20}\right)\right)\\ & =& {\cos }^{-1}\left(-\cos \left(\frac{3\mathrm{\pi }}{20}\right)\right)\mathbf{(}\mathbf{\because }\mathbf{cos}\left(\pi +\theta \right)\mathbf{=}\mathbf{-}\mathbf{cos}\left(\theta \right)\mathbf{)}\\ & =& \mathrm{\pi }-{\cos }^{-1}\left(\cos \left(\frac{3\mathrm{\pi }}{20}\right)\right)\mathbf{(}\mathbf{\because }\mathbf{}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\left(-x\right)\mathbf{=}\mathbf{\pi }\mathbf{-}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\left(x\right)\mathbf{)}\\ & =& \mathrm{\pi }-\frac{3\mathrm{\pi }}{20}\\ & =& \frac{17\mathrm{\pi }}{20}\\ & & \end{array}$

Hence, the principal value of ${\cos }^{-1}\left(\frac{1}{\sqrt{2}}\left[\cos \left(\frac{9\mathrm{\pi }}{10}\right)-\sin \left(\frac{9\mathrm{\pi }}{10}\right)\right]\right)$ is $\frac{17\mathrm{\pi }}{20}$.