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I have 3 functions of the same form, $x(p, q)$, $y(p, q)$, and $z(p, q)$ for which I am trying to evaluate $p$ and $q$. I know that $x+y+z=0$. I know I need to express p and q in terms of each other, but as you can see below, the three functions together are quite complex. I've been futzing around with trig identities to try and isolate either variable, but am stuck. Perhaps fresh eyes will help; can anyone here see a way forward?

All values are known, except $p$ and $q$.

$x(p,q) = [v_{1x} \cos(p)-u_{1x}\sin(p)][(C_{2x}+u_{2x}\cos(q)+v_{2x}\sin(q))-(C_{1x}+u_{1x}\cos(p)+v_{1x}\sin(p))]$

$y(p,q) = [v_{1y}\cos(p)-u_{1y}\sin(p)][(C_{2y}+u_{2y}\cos(q)+v_{2y}\sin(q))-(C_{1y}+u_{1y}\cos(p)+v_{1y}\sin(p))]$

$z(p,q) = [v_{1z} \cos(p)-u_{1z}\sin(p)][(C_{2z}+u_{2z}\cos(q)+v_{2z}\sin(q))-(C_{1z}+u_{1z}\cos(p)+v_{1z}\sin(p))]$

Syntax
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  • This question doesn't seem appreciably different from your previous question. Voting to close. – Blue May 31 '18 at 00:45
  • Can we just close the original? This new question simplifies the issue a great deal, as I've gained some perspective in the way forward. I feel the original has a confounding amount of detail and lacks visibility, making this one much more likely to lead to an answer/solution. – Syntax May 31 '18 at 01:02
  • Since the original question has no answers, but has comments that others might find useful, you should simply edit it with your streamlined discussion, un-delete it, and delete this one. You might keep some of that discussion, however; knowing that the $u$s and $v$s are components of unit vectors can be helpful for simplification purposes. That said, you're ultimately trying to solve for $q$ in $$A \cos q + B \sin q = C$$ (and hoping that the result reduces nicely ... which is unlikely). That kind of thing is addressed in this question. – Blue May 31 '18 at 02:55
  • Ok. Before I do that, can you explain a bit how you arrived at the form $A \cos q + B \sin q = C$? – Syntax May 31 '18 at 03:10
  • Your $x$ expression has the form $$A_x\cos q+B_x\sin q+C_x$$ where $$\begin{align}A_x&=(v_{1x}\cos p-u_{1x}\sin p)u_{2x}\ B_x&=(v_{1x}\cos p-u_{1x}\sin p)v_{2x}\ C_x&=(v_{1x}\cos p-u_{1 x}\sin p)(C_{2x}-(C_{1x}-u_{1x}\cos p-v_{1x}\sin p))\end{align}$$ Likewise for $y$ and $z$. Thus, $x+y+z=0$ amounts to $$(A_x+A_y+A_x)\cos q+(B_x+B_y+B_z)\sin q=-(C_x+C_y+C_z)$$ or, simply, $A\cos q + B \sin q = C$. That's the only equation that needs "solving", at which point you'll have $q$ in terms of $A$, $B$, $C$. Then, it's a matter of slogging-through to unpack those terms into $p$, $u$s, and $v$s. – Blue May 31 '18 at 03:38
  • @Blue I've spent some time trying to work out the intermediate steps to arrive at the definitions you've provided for $A_x$, $B_x$, and $C_x$. I can't work out what you've done to arrive there. Can you elaborate just a bit more? – Syntax Jun 30 '18 at 16:48
  • @ Blue Wouldn't it be the following? \begin{align}A_x&=(v_{1x}\cos p-u_{1x}\sin p)u_{2x}\ B_x&=(v_{1x}\cos p-u_{1x}\sin p)v_{2x}\ C_x&=(v_{1x}\cos p-u_{1 x}\sin p)(C_{2x}-C_{1x}-u_{1x}\cos p-v_{1x}\sin p)\end{align} – Syntax Jun 30 '18 at 17:04

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