Question

The principal value of ${\cos }^{-1}\left(-\frac{1}{2}\right)$ is equal to

Answer 1

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

As we know that, ${\cos }^{-1}\left(-x\right)=\mathrm{\pi }-{\cos }^{-1}\left(x\right)$

$\therefore {\cos }^{-1}\left(-\frac{1}{2}\right)=\mathrm{\pi }-{\cos }^{-1}\left(\frac{1}{2}\right).....\left(i\right)$

Now, Let, $\begin{array}{rcl}{\cos }^{-1}\left(\frac{1}{2}\right)& =& y\end{array}$

$\begin{array}{rcl}& \Rightarrow & \cos \left(y\right)=\frac{1}{2}\\ \cos \left(y\right)& =& \cos \left(\frac{\mathrm{\pi }}{3}\right)\\ & \Rightarrow & y=\frac{\mathrm{\pi }}{3}\end{array}$

So, $\begin{array}{rcl}{\cos }^{-1}\left(\frac{1}{2}\right)& =& \frac{\mathrm{\pi }}{3}\end{array}$

$\begin{array}{rcl}\therefore {\cos }^{-1}\left(-\frac{1}{2}\right)& =& \mathrm{\pi }-{\cos }^{-1}\left(\frac{1}{2}\right)(fromeq.(i\left)\right)\\ & =& \mathrm{\pi }-\frac{\mathrm{\pi }}{3}\\ & =& \frac{2\mathrm{\pi }}{3}\end{array}$

Hence, the principal value of ${\cos }^{-1}\left(-\frac{1}{2}\right)$ is $\frac{2\mathrm{\pi }}{3}$.