$\sqrt{\left( r_d \cos\left(\frac{-4(C – X) \csc(2α)}{Z}\right) + r_p \left(\frac{2C \tan(α) – 4 (C – X) \csc(2α))}{Z}\right) \sin\left(\frac{-4 (C – X) csc(2α)}{Z}\right) – m X \tan(α) \sin\left(\frac{-4 (C – X) \csc(2α)}{Z}\right)\right)^2 + \left( r_d \sin\left(\frac{4(C – X) \csc(2α)}{Z}\right) – r_p \left(\frac{2C \tan(α) – 4 (C – X) \csc(2α))}{Z}\right) \cos\left(\frac{-4 (C – X) csc(2α)}{Z}\right) + m X \tan(α) \cos\left(\frac{-4 (C – X) \csc(2α)}{Z}\right)\right)^2}$
- $r_d = r_p + m X – m C$
- $r_p = \frac {m Z}{2}$
- $m$ is positive
- $α$ is between $0$ and $\frac π 4$
- $Z$ is a positive integer
- $X$ is between -1 and +1
- $C$ is between 1 and 1.5
I've been staring at this until I'm cross-eyed, but I can't find any way to break it down. It's not for a class, so I don't have any resources to call upon.
Context:
I'm trying to find the radius of a point on the curve defined by the parametric expressions
$$x = r_d \cos(γ) + r_p \left(\frac{2C \tan(α)} Z + γ\right) \sin(γ) – m X \tan(α) \sin(γ),$$
$$y = r_d \sin(γ) – r_p \left(\frac{2C \tan(α)} Z + γ\right) \cos(γ) + m X \tan(α) \cos(γ)$$
Specifically, the point at $γ = \frac{-4(C – X) \csc(2α)}{Z}$. My instinct for solving that was to plug in the value and use the Pythagorean theorem, which created the expression that is the subject of this question. If there's a better way to find this radius, I would love to try it out.
Update: Looking to simplify the base expressions, I can expand the instances of $r_d$ and $r_p$ and then factor out the $m$ from all three terms, and I can factor out the $\sin$ and $\cos$ from the last two terms, but I can't see how to use that to any advantage…
Best Answer
It's often much better to simplify as soon as possible. In this case, the parameterized $x$ and $y$ values at the specified value $\gamma_0 := -4(C-X)\csc(2\alpha)/Z$ reduces fairly nicely:
$$\begin{align} x &=\tfrac12 mZ \left(\;\cos\gamma_0 + \gamma_0 \cos\alpha \sin(\alpha+\gamma_0)\;\right) \tag1\\[4pt] y &=\tfrac12 mZ \left(\;\sin\gamma_0 - \gamma_0 \cos\alpha \cos(\alpha+\gamma_0)\;\right) \tag2 \end{align}$$
From there, we easily get $$x^2+y^2 = \tfrac14m^2Z^2\left(\;1 + \gamma_0 \sin 2 \alpha + \gamma_0^2\cos^2\alpha\;\right) \tag3$$ (Conveniently, there are no $\gamma$s inside the trig functions.)
If you like, you can expand $1=\sin^2\alpha+\cos^2\alpha$ and $\sin2\alpha=2\sin\alpha\cos\alpha$, regroup, and write $$x^2+y^2 = \tfrac14m^2Z^2\left(\;\left(\gamma_0\cos\alpha+\sin\alpha\right)^2+\cos^2\alpha\;\right) \tag4$$
At this point, expanding $\gamma_0$ explicitly to $-4(C-X)\csc(2\alpha)/Z$ doesn't seem to give anything particularly pretty, so I'll leave that to the reader. $\square$
As a bit of a prequel, just substituting $r_d\to r_p+mX-mC$and $r_p\to mZ/2$ into OP's parametric equations gives the simplification $$\begin{align} x &= \tfrac12 mZ \left(\; \cos\gamma + \gamma \sin\gamma +\gamma_0 \sin\alpha\cos(\alpha+\gamma) \;\right) \tag{0.1}\\[4pt] y &= \tfrac12 mZ \left(\; \sin\gamma - \gamma \cos\gamma +\gamma_0 \sin\alpha \sin(\alpha+\gamma) \;\right) \tag{0.2} \end{align}$$ with $\gamma_0$ as above. From these, we get $$x^2 + y^2 = \tfrac14 m^2Z^2 \left(\; 1 + \gamma_0 \sin2\alpha + \gamma^2\cos^2\alpha + (\gamma-\gamma_0)^2 \sin^2\alpha \;\right) \tag{0.3}$$ When $\gamma=\gamma_0$, we have that $(0.1)$, $(0.2)$, $(0.3)$ reduce to $(1)$, $(2)$, $(3)$.