Question

How do you find ${\sin }^{-1}\left(-\frac{1}{2}\right)$?

Answer 1

Use the standard Trigonometric angles

We know that,$\sin 30°=\frac{1}{2}$

The given function can be solved as,

$$\begin{gathered} {\sin }^{-1}\left(-\frac{1}{2}\right) \\ ={\sin }^{-1}\left(\sin \left(\frac{11\pi }{6}\right)\right)\left[\because \sin \left(\frac{11\pi }{6}\right)=-\frac{1}{2}\right] \\ =\frac{11\pi }{6}\left[\because {\sin }^{-1}\left(\sin \left(x\right)\right)=x\right] \end{gathered}$$

Similarly

$$\begin{gathered} {\sin }^{-1}\left(-\frac{1}{2}\right) \\ ={\sin }^{-1}\left(\sin \left(\frac{7\pi }{6}\right)\right)\left[\because \sin \left(\frac{7\pi }{6}\right)=-\frac{1}{2}\right] \\ =\frac{7\pi }{6}\left[\because {\sin }^{-1}\left(\sin \left(x\right)\right)=x\right] \end{gathered}$$

Hence, the value of ${\sin }^{-1}\left(-\frac{1}{2}\right)$ is $\frac{7\pi }{6},\frac{11\pi }{6}$ for range $\left[0,2\pi \right]$