Question

Find the value of ${\tan }^{-1}\left(1\right)+{\cos }^{-1}(-\frac{1}{2})+{\sin }^{-1}(-\frac{1}{2}).$

Answer 1

Let $\tan y=1$
$\Rightarrow \tan y=\tan \left(\frac{\pi }{4}\right)$
We know that the range of the principal value branch of ${\tan }^{-1}$ is $(\frac{-\pi }{2},\frac{\pi }{2})$.
$\therefore y=\frac{\pi }{4}$
Hence, the principal value of ${\tan }^{-1}\left(1\right)$ is $\frac{\pi }{4}.$

Finding value of ${\cos }^{-1}(-\frac{1}{2})$
Let $y={\cos }^{-1}(-\frac{1}{2})$
$\Rightarrow \cos y=-\frac{1}{2}$
$\Rightarrow \cos y=\cos \left(\frac{2\pi }{3}\right)$
We know that the range of the principal value branch of ${\cos }^{-1}x$ is $[0,\pi ]$.
$\therefore y=\frac{2\pi }{3}$
Hence, the principal value of ${\cos }^{-1}(-\frac{1}{2})$ is $\frac{2\pi }{3}.$

Finding value of ${\sin }^{-1}(-\frac{1}{2})$
Let $y={\sin }^{-1}(-\frac{1}{2})$
$\Rightarrow \sin y=-\frac{1}{2}$
$\Rightarrow \sin y=\sin (-\frac{\pi }{6})$
We know that the range of the principal value branch of ${\sin }^{-1}x$ is $[-\frac{\pi }{2},\frac{\pi }{2}]$.
$\therefore y=-\frac{\pi }{6}$
Hence, the principal value of ${\sin }^{-1}(-\frac{1}{2})$ is $-\frac{\pi }{6}$

Finding the value of given expression
${\tan }^{-1}\left(1\right)+{\cos }^{-1}(-\frac{1}{2})+{\sin }^{-1}(-\frac{1}{2})=\frac{\pi }{4}+\frac{2\pi }{3}-\frac{\pi }{6}$

$=\frac{3\pi +8\pi -2\pi }{12}$
$=\frac{3\pi }{4}$