Prove that $\arcsin(-x)= -\arcsin x$
Is it mathematically permissible to prove this using semi unit circles?
For example I proved it this way:
Consider the veritcal semi unit circle between y coordinates $\dfrac{\pi}{2}$ and $-\dfrac{\pi}{2}$
For $x>0$, $\arcsin (x) = \theta $
From symmetry, $x$ is vertically above $-x$ and there are two congruent triangles with same base. Now, we just need to change the direction to measure the other angle which can be done by placing a minus sign before the first angle obtained.
Thus, $-\arcsin x= \arcsin -x$.
QED
Is this proof fine?
Similarly we can prove (using semi unit circles) identities like: $\arccos(-x)= \pi - \arccos(x)$
It looks good but a little vague. Instead of symmetry all you really need to rely on is the fact that $\sin$ is odd. The definition of the $\arcsin$ is that $\theta = \arcsin x$ if and only if $\theta \in [-\pi/2,\pi/2]$ and $\sin \theta = x.$ You get \begin{align*} \theta = \arcsin (-x) &\iff \theta \in [-\pi/2,\pi/2] \text{ and } \sin \theta = -x \\ &\iff - \theta \in [-\pi/2,\pi/2] \text{ and } \sin(-\theta) = x\\ &\iff - \theta = \arcsin x. \end{align*}
Thus $\theta = \arcsin(-x)$ and $-\theta = \arcsin x$ are logically equivalent. You can equate $\theta$ to get the identity you want.
ADDENDUM: Your question about $\arccos$ has a similar answer. Its definition is that $\theta = \arccos x$ if and only if $\theta \in [0,\pi]$ and $\cos \theta = x$. Following a similar list of implications using the identity $\cos (\pi - \theta) = \cos \pi \cos \theta + \sin \pi \sin \theta = -\cos \theta$ you get \begin{align*} \theta = \arccos(-x) &\iff \theta \in [0,\pi] \text{ and } \cos \theta = -x \\ &\iff - \theta \in [-\pi,0] \text{ and } -\cos(\theta) = x \\ &\iff \pi - \theta \in [0,\pi] \text{ and } \cos(\pi - \theta) = x\\ &\iff \pi - \theta = \arccos x. \end{align*} As above you can equate $\theta$ to find $\pi - \arccos(-x) = \arccos x$.