Check that the rotation isometry has an inverse The question was asked by my professor sort of abstractly but they want a mock proof so I'm trying to build upon my understanding of what he's asking:
Rotating a function is defined around $(0,0)$ by $\theta$.
the function $f:(x,y) \mapsto (x\cos\theta - y\sin\theta, x\sin\theta + y\cos\theta)$. 
I understand that for a function $g$ to be an inverse of $f$, $fg=i$ and $gf=i$ where $i$ is the identity function $i:(x.y)\mapsto (x,y)$.
We were given the hint to check $-\theta$ as $g$ but i don't fully understand how to implement that into actual arithmetic where the sines and cosines would cancel to be just $(x,y)$. can anybody help me find a starting point?
 A: Start from the hint $g:(x,y)\mapsto(x\cos(-\theta)-y\sin(-\theta),x\sin(-\theta)+y\cos(-\theta))$. Using the identities $\cos(-\theta)=\cos\theta$ and $\sin(-\theta)=-\sin\theta$ makes this $(x\cos\theta+y\sin\theta,-x\sin\theta+y\cos\theta)$. Now form $g\circ f(x,y)$ and simplify. The first component is $$(x\cos\theta-y\sin\theta)\cos\theta+(x\sin\theta+y\cos\theta)\sin\theta = (\cos^2\theta+\sin^2\theta)\,x+(\cos\theta\sin\theta-\cos\theta\sin\theta)\,y = x.$$ For the second, we have $$(x\sin\theta+y\cos\theta)\cos\theta-(x\cos\theta-y\sin\theta)\sin\theta = (\cos\theta\sin\theta-\cos\theta\sin\theta)\,x + (\cos^2+\sin^2)\,y = y.$$
A: Hint:  replace $\theta$ by $-\theta$ ...
...in the matrix:
 $\begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{matrix}$
And, use that $\cos(-\theta) =\cos (\theta)$ and $\sin (-\theta)= -\sin(\theta)$, that is $\cos$ is even and $\sin$ is odd...
If you apply one matrix followed by the other to the column vector  $\bigg (\begin{matrix}x \\ y \end{matrix}\bigg) $  you will of course get $\bigg (\begin{matrix}x \\ y \end{matrix}\bigg) $ back...  because rotation by $\theta$  followed by $-\theta$ is the identity...

the inverse is $(x,y)\to ((\cos\theta)x+(\sin\theta)y, -(\sin\theta)x+(\cos\theta)y)$

