Solving for $y_3$ in a complicated expression

Question

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 $z_2$?

Answer 1

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 $c=z_1+z_2$

Answer 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.$\qquad\square$

Answer 3

If $\tan\theta = \frac{p}{q}$, with positive $q$, then $\theta$ is a First- or Fourth-Quadrant angle, and $\sin\theta = \frac{p}{\sqrt{p^2+q^2}}$.

For $$\theta = \operatorname{atan}\frac{y_3-z_1}{\sqrt{(z_1+z_2)^2-(y_3-z_1)^2}}$$ we have $$p = y_3 - z_1, \qquad q = \sqrt{(z_1+z_2)^2 - (y_3-z_1)^2}, \qquad \sqrt{p^2+q^2} = \sqrt{(z_1+z_2)^2}=|z_1+z_2|$$ so that $$\begin{align} y_3 = z_1 + (z_1+z_2)\sin\theta \quad&\implies\quad y_3=z_1+\frac{z_1+z_2}{|z_1+z_2|}(y_3-z_1)= z_1 \pm ( y_3 - z_1 )\\ &\implies\quad y_3=z_1+ y_3 - z_1\quad\text{or}\quad y_3=z_1-y_3+z_1 \\[6pt] &\implies\quad 0 = 0 \quad\text{or}\quad y_3 = z_1 \end{align}$$ Consequently, $$\begin{align} z_1+z_2 > 0 \quad&\implies\quad y_3 \in \mathbb{R} \\ z_1+z_2 < 0 \quad&\implies\quad y_3 = z_1 \\ z_1+z_2 = 0 \quad&\text{undefined (why?)} \end{align}$$