Solving $\sin \theta_1+\sin \theta_2=\sin \theta_3+c_1$ and $\cos \theta_1+\cos \theta_2=\cos\theta_3+c_2$ for $\theta_1$ and $\theta_2$

Question

For these two equations: $$\begin{align} \sin \theta_1+\sin \theta_2&=\sin \theta_3+c_1 \\ \cos \theta_1+\cos \theta_2&=\cos\theta_3+c_2 \end{align}$$ where $c_1$ and $c_2$ are constants.

How can I put $\theta_1$ as a function of $\theta_3$ only, and $\theta_2$ as a function of $\theta_3$ only?

Answer 1

Of course, there are algebraic ways to solve this. But this problem is easier to deal with geometrically:

If $B=(x,y)=(c_2+\cos\theta_3,c_1+\sin\theta_3)$, then it's easy to see that: $$ \frac{\theta_1+\theta_2}2 = \arg B=\mathop{\mathrm{sign}}y \ \arccos\frac{x}{\sqrt{x^2+y^2}},\\ \cos{\frac{\theta_2-\theta_1}2} = \frac{|B|}2=\frac{\sqrt{x^2+y^2}}2 $$

This assumes, that $-\pi<\frac{\theta_1+\theta_2}2<\pi$ and $0\le\theta_2-\theta_1\le\pi$. If you need other assumptions, you can modify the result accordingly.