Question

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

Answer 1

Solve using standard trigonometric angle value:

The given inverse trigonometric function is ${\cos }^{-1}\left(-\frac{1}{2}\right)$.

We know, that,

${\cos }^{-1}\left(-\frac{1}{2}\right)={\cos }^{-1}\left(-\cos \frac{\mathrm{\pi }}{3}\right)$ $\left[\because \cos \frac{\pi }{3}=\frac{1}{2}\right]$

$\Rightarrow$ ${\cos }^{-1}\left(-\frac{1}{2}\right)={\cos }^{-1}\left[\cos \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)\right]$ $[\because \cos \left(\pi -x\right)=-\cos x]$

$\Rightarrow$ ${\cos }^{-1}\left(-\frac{1}{2}\right)=\left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)$ $\left[\because {\cos }^{-1}\left(\cos n\right)=n\right]$

$\Rightarrow$ ${\cos }^{-1}\left(-\frac{1}{2}\right)=\frac{3\mathrm{\pi }-\mathrm{\pi }}{3}$

$\Rightarrow$ ${\cos }^{-1}\left(-\frac{1}{2}\right)=\frac{2\mathrm{\pi }}{3}$

Hence, the required value is $\frac{2\mathrm{\pi }}{3}$.