Solving a system of two trigonometric equations

Question

I have to solve the following system made of two equations.

• The variables are $x_i$ for $i=1,...,n$.
• For the parameters, we have $a_i\in\mathbb{R}$ and $B\geq0$.

The two equations are: $$\sum_{i=1}^na_i\cos(x_i)=B,$$ $$\sum_{i=1}^na_i\sin(x_i)=0.$$

I see a condition of existence: we must have $B\leq \sum_{i=1}^n|a_i|$.

But are there any other conditions? And my main question is: how could I find a solution to such equations?

Thanks for you help!

Answer 1

We can reformulate the problem using complex exponentiation

$$B - \sum_{i=1}^n a_i s_i = 0$$

$$s_k = \exp(i x_k)$$

And interpret it as a polygon space of varied angles in vertices. It doesn't look like a simple problem. This may be of use:

http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.237.2549

(You might also want to ask it on MathOverflow.)

Answer 2

This describes an articulated arm with section lengths $a_i$.

With a single section, you reach the circle of radius $a_0$.

With two sections, you can sweep the above circle by moving its center on the circle of radius $a_1$. Then you can reach any point in the ring of radii $r_1=|a_0-a_1|$ and $R_1=a_0+a_1$. For such a point, there are two solutions.

With a third section and more, you can sweep the above ring and reach the points between the radii $r_k=r_{k-1}-a_k$ and $R_k=R_{k-1}+a_k$. For the inner radius, if some permutation of the sections leads to a negative value, the whole disk can be reached. For these points, there is a multiple infinity ($k-2$) of solutions.