It is not a school-related problem, and pretty much the title. I am looking for a way to solve the system of trigonometric equations:
$$
A\cos(x) + B\cos(y) = C \tag{1}
$$
$$
A\sin(x) + B\sin(y) = D \tag{2}
$$
I have already solved the system for the specific case of $A=B$ and am looking to generalize this to $A \neq B$.
For reference, my solution to the specific case $A=B$ is:
$$x = \cos^{-1}(\frac{C}{A} – \cos(\cot^{-1}(\frac{C}{D})-\cos^{-1}(\frac{D}{2A\sin(\cot(\frac{C}{D})})) \tag{3}$$
$$y = \cot(\frac{C}{D}) – \cos^{-1}(\frac{D}{2A\sin(\cot(\frac{C}{D})})) \tag{4}$$
I have been using computer software such as WolframAlpha to solve it, and they do solve it, but I've been unable to come up with a solution by hand after a couple months of working on it pretty regularly.
Can it be solved by hand?
Best Answer
Move either the terms with $A$ or $B$ to the other side and square: $$B^2\cos^2y=A^2\cos^2x-2AC\cos x+C^2\\B^2\sin^2y=A^2\sin^2x-2AD\sin x+D^2$$
Now add them together: $$B^2=A^2+C^2+D^2-2AC\cos x-2AD\sin x$$
Now write $C=R\cos\phi$, $D=R\sin\phi$. So $R^2=C^2+D^2$. $$B^2-A^2-C^2-D^2=-2A\sqrt{C^2+D^2}(\cos x\cos\phi+\sin x\sin\phi)\\\cos(x-\phi)=\frac{-B^2+A^2+C^2+D^2}{2A\sqrt{C^2+D^2}}$$
We take inverse to get $x$.
Likewise , we can get $y$.
Can you take it from here?