Solving for $\alpha$ and $\beta$ in the system $x=d\cos(\alpha+\beta)+h\cos\alpha$, $y=d\sin(\alpha+\beta)+h\sin\alpha$ How do I solve the following trigonometric equations for $\alpha$ and $\beta$:
$$ x = d\cos(\alpha + \beta)+h\cos(\alpha) $$
$$ y = d\sin(\alpha + \beta)+h\sin(\alpha) $$
My attempt:
Use:
$$ \sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$$
$$ \cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$$
So:
$$ x = d\cos(\alpha + \beta)+h\cos(\alpha) \implies d(\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta))+h\cos(\alpha)=x$$
$$ y = d\sin(\alpha + \beta)+h\sin(\alpha)  \implies d(\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta))+h\sin(\alpha) = y$$
Solve the first equation for $\beta$:
$$\frac{x-h\cos(\alpha)}{d} =  \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) \implies ??\text{ (I'm stuck here) }$$
I would be very grateful for any hints about how to continue from there.
 A: After squaring you have
\begin{align}
x^2&=d^2\cos^2(\alpha+\beta)+2dh\cos(\alpha+\beta)\cos\alpha+h^2\cos^2\alpha\\
y^2&=d^2\sin^2(\alpha+\beta)+2dh\sin(\alpha+\beta)\sin\alpha+h^2\sin^2\alpha
\end{align}
Summing them up gives
$$
2dh(\cos(\alpha+\beta)\cos\alpha+\sin(\alpha+\beta)\sin\alpha)
=x^2+y^2-d^2-h^2
$$
The left hand side is clearly $\cos\beta$, so we have got
$$
\cos\beta=\frac{x^2+y^2-d^2-h^2}{2dh}
$$
Expanding the equations with the addition formulas gives
$$
\begin{cases}
(d\cos\beta+h)\cos\alpha-d\sin\beta\sin\alpha=x \\[6px]
d\sin\beta\cos\alpha+(d\cos\beta+h)\sin\alpha=y
\end{cases}
$$
and Cramer's rule provides
$$
\begin{cases}
\cos\alpha=
\dfrac
  {x(d\cos\beta+h)+dy\sin\beta}
  {(d\cos\beta+h)^2+d^2\sin^2\beta}
=
\dfrac
  {x(d\cos\beta+h)+dy\sin\beta}
  {x^2+y^2}
\\[8px]
\sin\alpha=
\dfrac
  {y(d\cos\beta+h)+dx\sin\beta}
  {(d\cos\beta+h)^2+d^2\sin^2\beta}
=
\dfrac
  {y(d\cos\beta+h)+dx\sin\beta}
  {x^2+y^2}
\end{cases}
$$
Since $\sin\beta$ is determined (up to the sign) by $\cos\beta$, you have the requested formulas.
Note that the denominators can be rewritten
$$
d^2\cos^2\beta+2dh\cos\beta+h^2+d^2\sin^2\beta=
d^2+h^2+2dh\cos\beta=
x^2+y^2
$$
A: Mathematica finds that one of the solutions is:
$\alpha = \frac{h x \left(-d^2+h^2+x^2+y^2\right)-\sqrt{-h^2 y^2 \left((d-h)^2-x^2-y^2\right) \left((d+h)^2-x^2-y^2\right)}}{h^2
   \left(x^2+y^2\right)}$
and
$\beta = \tan ^{-1}\left(\frac{-d^2-h^2+x^2+y^2}{2 d h}\right)$.
A: $$\frac{x-h\cos(\alpha)}{d}=\cos(\alpha+\beta)$$
$$\alpha+\beta=\cos^{-1}\left(\frac{x-h\cos(\alpha)}d\right)$$
In the second equation, we have:
$$\frac{y-h\sin(\alpha)}{d}=\sin(\alpha+\beta)$$
$$\alpha+\beta=\sin^{-1}\left(\frac{y-h\sin(\alpha)}d\right)$$
Thus, we have:
$$\cos^{-1}\left(\frac{x-h\cos(\alpha)}d\right)=\sin^{-1}\left(\frac{y-h\sin(\alpha)}d\right)$$
We know that $\cos(\sin^{-1}(\mu))=\sqrt{1-\mu^2}$.
$$\frac{x-h\cos(\alpha)}d=\sqrt{1-\left(\frac{y-h\sin(\alpha)}d\right)^2}$$
$$\left(\frac{x-h\cos(\alpha)}d\right)^2=1-\left(\frac{y-h\sin(\alpha)}d\right)^2$$
$$(x-h\cos(\alpha))^2=d^2-(y-h\sin(\alpha))^2$$
$$x^2+y^2-2hx\cos(\alpha)-2hy\sin(\alpha)+h^2\cos^2(\alpha)+h^2\sin^2(\alpha)=d^2$$
$$-2h(x\cos(\alpha)+y\sin(\alpha))+h^2(\cos^2(\alpha)+\sin^2(\alpha))=d^2-x^2-y^2$$
$$-2h(x\cos(\alpha)+y\sin(\alpha))+h^2(1)))=d^2-x^2-y^2$$
$$x\cos(\alpha)+y\sin(\alpha)=\frac{x^2+y^2+h^2-d^2}{2h}$$
This is as close as I can take you, hopefully someone can build upon this?
