Solving for $y_3$ in a complicated expression I need to solve this equation for $y_3$:

$$y_3=z_1+(z_1+z_2)\sin\left(\arctan\left(\frac{y_3-z_1}{\sqrt{(z_1+z_2)^2–(y_3-z_1)^2}}\right)\right)$$

Is there a solution to this equation in terms of $z_1$ and $z_2$?
Consider the function
$$
f(x)=z_1+(z_1+z_2)\sin\left(\arctan\frac{x-z_1}{\sqrt{(z_1+z_2)^2-(x-z_1)^2}}\right)
$$
which is defined for $x\in I=(z_1-|z_1+z_2|,z_1+|z_1+z_2|)$. Calculating the derivative we see immediately that $f'(x)=1$ for every $x\in I$. So $f(x)=x+c$ for some constant $c$. But
$f(z_1)=z_1$, thus $f(x)=x$ for every $x\in I$. Thus the interval 
$$I=(z_1-|z_1+z_2|,z_1+|z_1+z_2|)$$
represents the set of solutions to this equation.$\qquad\square$
Let $\displaystyle\arcsin\frac bc=\theta$

$\displaystyle\implies\sin\theta=\frac bc$ and $\displaystyle-\frac\pi2\le\theta\le\frac\pi2$

$\displaystyle\implies\cos\theta\ge0\implies\cos\theta=+\frac{\sqrt{c^2-b^2}}c$

$\displaystyle\implies\tan\theta=+\frac b{\sqrt{c^2-b^2}}\implies\arctan \frac b{\sqrt{c^2-b^2}}=\theta=\arcsin\frac bc$

Here $b=y_3-z_1$ and $c=z_1+z_2$