Solve for $a$ and $b$ in $\begin{cases}x = l_1\times \sin(a) - l_2\times \sin(b)\\ y = l_1\times \cos(a) + l_2\times \cos(b)\end{cases}$.

Question

I have two equations and need to solve for two variables. I need an equation where no matter what $x$ or $y$ I pick that I can obtain what $a$ and $b$ should be. I have tried simulators online and using trig identities to solve for the variables.

$$\begin{cases}x = l_1\times \sin(a) - l_2\times \sin(b)\\ y = l_1\times \cos(a) + l_2\times \cos(b)\end{cases}$$

$l_1 = 4.5\\ l_2 = 6.5$

$x$ and $y$ can be coordinate. I have been using coordinates such as $(6.5, 4.5)$, $(11, 0)$, and $(0, 11)$

Thanks in advance!

Answer 1

writing $$x+L_2\sin(b)=L_1\sin(a)$$ $$y+L_2\cos(b)=L_1\cos(a)$$ squaring and adding we obtain $$x^2+y^2+L_2^2+2xL_2\sin(b)+2yL_2\cos(b)=L_1^2$$ and the we have $$x\sin(b)+y\cos(b)=\frac{L_1^2-L_2^2-x^2-y^2}{2L_2}$$ then we can write $$y\cos(b)=A-x\sin(b)$$ squaring again we get $$y^2\cos^2(b)=A^2+x^2\sin^2(b)-2Ax\sin(b)$$ with $$\cos^2(b)=1-\sin^2(b)$$ you will get a quadratic equation , solve this! Good luck!

Answer 2

There are two solutions. As I don't know what you have to do with them I wrote both

Hope this can help

$$ \begin{array}{ll} a=\arcsin \frac{x^3+\left(y^2-22\right) x-\sqrt{-y^2 \left(x^2+y^2-121\right) \left(x^2+y^2-4\right)}}{9 \left(x^2+y^2\right)} & b=-\arcsin\frac{x^3+\left(y^2+22\right) x+\sqrt{-y^2 \left(x^2+y^2-121\right) \left(x^2+y^2-4\right)}}{13 \left(x^2+y^2\right)} \\ a=\arcsin\frac{x^3+\left(y^2-22\right) x+\sqrt{-y^2 \left(x^2+y^2-121\right) \left(x^2+y^2-4\right)}}{9 \left(x^2+y^2\right)} & b=\arcsin \frac{\sqrt{-y^2 \left(x^2+y^2-121\right) \left(x^2+y^2-4\right)}-x \left(x^2+y^2+22\right)}{13 \left(x^2+y^2\right)} \\ \end{array} $$