Question

Find the principal value of ${{\cos }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)$

Answer 1

Hint: let us consider the y as given inverse trigonometric function and then we will get the value of $\cos y$ and if $\cos y$ is negative then the principal value will be $\pi -\theta$. The principal value of $\cos \theta$ lies between 0 and $\pi$

Complete step-by-step answer:
Let y=${{\cos }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
$\Rightarrow \cos y=\dfrac{-\sqrt{3}}{2}$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Since $\dfrac{-\sqrt{3}}{2}$is negative, principal value of $\theta$ is $\pi -\theta$
$\Rightarrow \cos y=\dfrac{-\sqrt{3}}{2}=\cos \left( \pi -\dfrac{\pi }{6} \right)$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
We know the principal value of ${{\cos }^{-1}}$is $\left[ 0,\pi \right]$
Hence the principal value of ${{\cos }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)$is
$\pi -\dfrac{\pi }{6}=\dfrac{6\pi -\pi }{6}=\dfrac{5\pi }{6}$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)

Note: The general formula for principal value of ${{\cos }^{-1}}\left( \sin \theta \right)=\dfrac{\pi }{2}-\theta$ if and only if $\theta \in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. The principal value of $\theta$ for any given inverse function should lie within its range. The range of inverse cos function or arc cos function is $\left[ 0,\pi \right]$. So the principle of $\theta$ for inverse cosine function always lies between 0 and $\pi$.