Question

Find the principal values of ${\sin }^{-1}\frac{1}{2}$

Answer 1

We know that ${\sin }^{-1}x$ denotes an angle in an interval $[\frac{-\pi }{2},\frac{\pi }{2}]$ whose sine is $x$ for $x\in [-1,1]$

$\therefore {\sin }^{-1}\left(\frac{1}{2}\right)=$ an angle in $[\frac{-\pi }{2},\frac{\pi }{2}]$ whose sine is $\frac{1}{2}$

$\Rightarrow {\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$