How do I simplify $\arccos(x)−\arcsin(x)$ for $x$ in $(−1,1)$ 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
 A: $\arccos(x)$ is the angle $\theta$ with $\cos(\theta) = x$ and $0 \le \theta \le \pi$, while $\arcsin(x)$ is the angle $\phi$ with $\sin(\phi) = x$ and $-\pi/2 \le \phi \le \pi/2$.  Now  $\sin(\theta) = \cos(\pi/2 - \theta)$ and $-\pi/2 \le \pi/2 - \theta \le \pi/2$, so $\pi/2 - \theta = \phi$.  Thus $$\arccos(x) - \arcsin(x) = \theta - \phi = 2\theta - \pi/2 = 2 \arccos(x) - \pi/2$$
or if you prefer
$$ \pi/2 - 2 \phi = \pi/2 - 2 \arcsin(x)$$
A: 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 $$\arccos(x)-\left(\frac{\pi}{2}-\arccos(x)\right)=2\arccos(x)-\frac{\pi}{2}$$
A: Setting
$$
u(x)=\arccos x-\arcsin x \quad\forall x \in (-1,1),
$$
we have
$$
u'(x)=-\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=-\frac{2}{\sqrt{1-x^2}}\quad\forall x \in (-1,1).
$$
It follows that
$$
\arccos x-\arcsin x=u(0)+\int_0^xu'(t)\,dt=\frac\pi2-2\int_0^x\frac{1}{\sqrt{1-t^2}}\,dt,
$$
i.e. for every $x\in (-1,1)$ we have:
$$
\arccos x-\arcsin x=\frac\pi2-2\arcsin x=\frac\pi2+2(\arccos x-\frac\pi2)=2\arccos x-\frac\pi2.
$$
