If anybody wants to have fun solving trigonometrics systems, I am highy interested. :)
I've got the following system to solve, which includes trigonometric functions.
$x_{0}$, $x_{1}$, $y_{0}$, and $y_{1}$, are known.
I'm looking for $\theta_{0}$ and $\theta_{1}$ and I am stuck... Any idea ?
$$ \begin{align}[x_{1}\cos(\theta_{1}) - y_{1}\sin(\theta_{1}) ] - [x_{0}\cos(\theta_{0}) - y_{0}\sin(\theta_{0})] &= (x_{1}-x_{0})^2 + (y_{1}-y_{0})^2 \\ [x_{1}\sin(\theta_{1}) + y_{1}\cos(\theta_{1}) ] - [x_{0}\sin(\theta_{0}) + y_{0}\cos(\theta_{0})] &= 0\end{align} $$
Any help will be gratefully accepted!
Thank you in advance :)
Let $x_i = r_i\cos\alpha_i$ and $y_i = r_i\sin\alpha_i$, the two equations becomes $$r_1\cos(\alpha_1+\theta_1) - r_0\cos(\alpha_0+\theta_0) = (x_1-x_0)^2+(y_1-y_0)^2$$ $$r_1\sin(\alpha_1+\theta_1) = r_0\sin(\alpha_0+\theta_0).$$ The second equation can further lead to $$r_1^2\cos^2(\alpha_1+\theta_1) = (r_1^2-r_0^2)+r_0^2\cos^2(\alpha_0+\theta_0).$$ Then you can eliminate $\cos(\alpha_1+\theta_1)$ from the first equation and solving a quadratic equation of $\cos(\alpha_0+\theta_0)$.