Negative inverse trig functions I was practising my integration earlier and for one question I got $-\arccos(x - 2)$, while the book has $\arccos(2 - x)$.
The answers seems so close and I was wondering (a) am I right? But more importantly, (b) is there some relationship between $-\arccos(-x)$ and $\arccos x$. More generally, is there a relationship between inverse trig functions and their negatives and what is it?
 A: $\arccos(-x)=\pi-\arccos x$, so you are correct.
You also have $\arcsin(-x)=-\arcsin x$, and $\arctan(-x)=-\arctan x$.

We can use the identity $\arccos x=\frac{\pi}{2}-\arcsin x$ to derive the first identity from the second, or vice versa,  and to show that the answer to the original problem could also have been written as $\arcsin(x-2)+C$. 
A: We have the identity $$\arccos(x)+\arccos(-x)=\pi\quad*$$
So $$-\arccos(x-2)=\arccos(2-x)-\pi$$
Since you've found the antiderivative, you have a function plus some constant of integration $C$. In this case, $C=-\pi$. So yes, you're right.
*$\quad x\in[-1,1]$
A: To see if $-\arccos{(x-2)}=\arccos{(2-x)}$, check the value at $x=2$. 
The first expression gives $-\arccos{(2-2)}=-\arccos{(0)}=-\pi/2$ 
And the second gives $\arccos{(2-2)}=\arccos{(0)}=\pi/2$
They must be different.
A: Inorder to understand the relationship between $\text{trig}^{-1}(x)$ and $\text{trig}^{-1}(-x)$ where $\text{trig}(x)$ is any fundamental trigonometric function, one must first come to terms with the relationship between $\text{trig(x)}$ and $\text{trig}(-x)$ which is as follows:
$$\sin(-x) = -\sin(x)\\
\cos(-x) = \cos(x)\\
\tan(-x) = -\tan(x)$$
The above is a direct result of the trig functions being either even or odd functions.
For an even function (where graph is symmetric about the y-axis), $f(-x) = f(x)$ while for an odd function (where graph is symmetric about the origin): $f(-x) = -f(x)$ 
If you are to draw the inverse of the graphs of the above function (Just constrict to suitable domain and reflect about $y=x$ line)http://www.purplemath.com/modules/invrsfcn.htm
Then, hopefully the relationships will be obvious from the graphs.
