If $f(x)=\sin^{-1}x +\cos^{-1} x$ and $g(x)$ are identical, find $g(x)$

Question

The given answer for this question is $|\sin^{-1}x| +\cos^{-1} |x|$.

I am completed fine with the answer, but why can’t $\sin^{-1} |x| +|\cos^{-1} x|$ also be right? It seems to satisfy all the necessary conditions. What am I missing?

Answer 1

Because, for instance,$$\arcsin\left(\left|-\frac12\right|\right)+\left|\arccos\left(-\frac12\right)\right|\ne\arcsin\left(-\frac12\right)+\arccos\left(-\frac12\right),$$since $\arcsin\left(\left|-\frac12\right|\right)\ne\arcsin\left(-\frac12\right)$, whereas $\left|\arccos\left(-\frac12\right)\right|=\arccos\left(-\frac12\right)$.

Answer 2

Note that $f(x) =π/2\ \forall\ x\in[-1,1]$.

For your function $$g(x)=\begin{cases} \sin^{-1}x+\cos^{-1}x=π/2 &,x\ge 0\\ \sin^{-1}(-x)+\cos^{-1}x &,x<0 \end{cases}$$

Hence, for $x<0$, $g(x) \in\left(\frac{π}{2},\frac{3π}{2}\right]$.