How do i simplify $\arccos(x)−\arcsin(x)$ for $x$ in $(−1,1)$
i got somewhere that…
$\sin(x)= \cos(\frac{\pi}{2}-x)$
so $\arccos(\sin(x))+x=\frac{\pi}{2}$
substituting that $\sin(x)=t \rightarrow \arcsin(t)=x$
$\arccos(t)+\arcsin(t)=\frac{\pi}{2}$
so working backwards
$\arccos(t)-\frac{\pi}{2} = -(\arcsin(t))$
hence $\arccos(x)−\arcsin(x) = \arccos(t) + \arccos(t)-\frac{\pi}{2} = 2\arccos(t)-\frac{\pi}{2}$
Which hasn't really simplified anything
Best Answer
This isn't a different conclusion from yours, but it is a different route. $\arccos(x)+\arcsin(x)$ is a constant $\frac{\pi}{2}$, which can be seen geometrically or with calculus. So then you have $$\arccos(x)-\left(\frac{\pi}{2}-\arccos(x)\right)=2\arccos(x)-\frac{\pi}{2}$$