$2$ by $2$ equation system where the desired variables are inside trigonometrics functions In this equation system, $X$ and $Y$ are known, I need to have $\alpha$ and $\beta$ in terms of $X$ and $Y$ (if I change $X$ and $Y$, what values do $\alpha$ and $\beta$ have?).
$$
X=60.5*\cos(  \beta  )-60*\cos(  \alpha  )
\\
Y=60.5*\sin(  \beta  )-60*\sin(  \alpha  )
$$
I started by getting $\beta$ from the first formula:
$$
  \beta =\arccos\left(\frac{\mathrm{X+60*\cos( \alpha )} }{\mathrm{60.5}}\right) 
$$
But when you substitute $\beta$ in the second equations things get complicated very fast:
$$
  Y=60.5*\sin\left(  \arccos\left(\frac{\mathrm{X+60*cos( \alpha )} }{\mathrm{60.5}}\right)  \right)-60*\sin(  \alpha  )
$$
I can apply the identity $  \sin(\arccos(x))= \sqrt{1-x^2} $ but that just keep complicating the equation.
Am I doing this correctly? Is it that complicated?
EDIT
Following the suggested answers, by squaring the equation an adding them up and applying some trigonometric identities, I get this expression:
$$
 \alpha  -  \beta =  \arccos ( \frac{\mathrm{7260.25-X^2-Y^2} }{\mathrm{7260}} )
$$
Wich is the cleanest I got so far. But then again, I'm stuck, what can I do now? Are there any other clever manipulations I can do to reduce the system and get another expression of alpha and beta?
 A: As you did, squaring and adding, you end with
$$X^2+Y^2=\frac{29041}{4}-7260 \cos (\alpha -\beta )$$ Solve for $\beta$ to get
$$\beta=\alpha +\epsilon \cos ^{-1}\left(\frac{29041-4(X^2+Y^2)}{29040}\right)\qquad (\epsilon=\pm 1)$$
Using $\epsilon=+ 1$, plug in the first equation and expand the cosine to get
$$480 X= \left(241-4(X^2+Y^2)\right)\cos (\alpha )-29040 \sqrt{1-\frac{\left( 29041-4(X^2+Y^2)\right)^2}{843321600}}\sin(\alpha )$$ which is a basic trigonometric equation in $\alpha$.
A: Square and add the two equations. You will get an expression for $\alpha-\beta$. Then you can use this. So, yeah. It comes out pretty complicated.
A: I would define 
$$
F(\alpha,\beta) = \Big(
X-60.5\cos\beta + 60\cos\alpha
\Big)^2+\Big(
Y-60.5\sin\beta + 60 \sin\alpha
\Big)^2 \geq 0
$$
and then numerically minimize $F(\alpha,\beta)$ globally over $(\alpha,\beta)\in [0,2\pi]\times [0,2\pi]$. Let $(\alpha^*,\beta^*)$ be the locations of the minima. If $F(\alpha^*,\beta^*)>0$ for all minima then the system does not have a solution. If $F(\alpha^*,\beta^*)=0$ then you have your solution.
