Pair of simultaneous trigonometric equations

Question

I have reduced a set of simultaneous equations to the following two equations. They represent the intersection between an ellipse (LHS) and a polynomial line (RHS). I can't think of a way to reduce it so it is solvable. It may, in fact, be unsolvable, but I want to make sure.

Thank you for any help in advance.

$$ -a.cos\theta = R_1.cos(\alpha-\beta_1)+C_1 $$ $$ -b.sin\theta = R_2.cos(\alpha-\beta_2)+C_2 $$

Answer 1

I presume you want to solve for $\theta$ and $\alpha$. Converting to complex exponentials with $\exp(i \theta) = w$, $\exp(i\alpha) = z$, $\exp(i \beta_1) = u_1$, $\exp(i \beta_2) = u_2$, the equations become $$ \eqalign{- \frac{a}{2} \left(w + \frac{1}{w}\right) &= \frac{R_1}{2} \left( \frac{z}{u_1} + \frac{u_1}{z}\right) + C_1\cr \frac{ib}{2} \left(w - \frac{1}{w}\right) &= \frac{R_2}{2} \left( \frac{z}{u_2} + \frac{u_2}{z}\right) + C_2\cr} $$ Putting them over a common denominator and taking the resultant with respect to $w$, we get a quartic equation for $z$:

$$ \left( {a}^{2}{R_{{2}}}^{2}{u_{{1}}}^{2}+{b}^{2}{R_{{1}}}^{2}{u_{{2}} }^{2} \right) {z}^{4}+4\,u_{{1}}u_{{2}} \left( {a}^{2}C_{{2}}R_{{2}}u_ {{1}}+{b}^{2}C_{{1}}R_{{1}}u_{{2}} \right) {z}^{3}+2\,{u_{{1}}}^{2}{u_ {{2}}}^{2} \left( -2\,{a}^{2}{b}^{2}+2\,{a}^{2}{C_{{2}}}^{2}+{a}^{2}{R _{{2}}}^{2}+2\,{b}^{2}{C_{{1}}}^{2}+{b}^{2}{R_{{1}}}^{2} \right) {z}^{ 2}+4\,{u_{{1}}}^{2}{u_{{2}}}^{2} \left( {a}^{2}C_{{2}}R_{{2}}u_{{2}}+{ b}^{2}C_{{1}}R_{{1}}u_{{1}} \right) z+{R_{{1}}}^{2}{b}^{2}{u_{{1}}}^{4 }{u_{{2}}}^{2}+{R_{{2}}}^{2}{a}^{2}{u_{{1}}}^{2}{u_{{2}}}^{4} = 0 $$ This has a "closed-form" solution in radicals, but it's not likely to be pretty.

Answer 2

Assuming, as Robert Israel did, that you want to solve for $\theta$ and $\alpha$, you can reduce the problem to a single equation $$-a\cos(\theta) = R_1 \cos(\alpha-\beta_1)+C_1\implies \cos(\theta)=-\frac{R_1 \cos(\alpha-\beta_1)+C_1}a$$ $$-b\sin(\theta) = R_2 \cos(\alpha-\beta_2)+C_2\implies \sin(\theta)=-\frac{R_2 \cos(\alpha-\beta_2)+C_2 } b$$ Now, using $\sin^2(\theta)+\cos^2(\theta)=1$, we end with the problem of finding the zero(s) of function $$F(\alpha)=b^2\left( R_1 \cos(\alpha-\beta_1)+C_1 \right)^2+a^2\left( R_2 \cos(\alpha-\beta_2)+C_2\right)^2-a^2b^2$$ For a given set of $a,b,\beta_1,\beta_2,R_1,R_2,C_1,C_2$ graphing the function should show the regions where the roots are located. Now, use Newton method to polish the solutions. When $\alpha$ is know, go back to the original equation to get the corresponding $\theta$'s.