I would like to minimize an angle based on two rational numbers $a,b$. The trigonometric equations read,
$ \cos\theta = \left(a-\frac{b}{2}\right)-\cos(1.6^\circ)\left(a-\frac{b}{2}\right)+\frac{\sqrt{3}}{2}b\sin\left(1.6^\circ\right)+1 $
and,
$ \sin\theta = \frac{\sqrt{3}}{2}b\left(1-\cos(1.6^\circ)\right)-\sin(1.6^\circ)\left(a-\frac{b}{2}\right). $
The only constraint that I have for the angle $\theta$ is that it's within the range $-5 \leq \theta \leq 5$. Someone mentioned how I could use a Jacobian matrix to minimize the angle, but I am not too familiar with that procedure. Is there another way to minimize the angle, and thus solving for the set of equations?
EDIT: I have read upon other similar problems and made a bit of progress but I am not sure if I am going in the right direction. I solved for theta in the first equation,
$ \theta = \arccos \left( \left(a - \frac{b}{2} \right)(1-\cos(1.6^\circ)) + \frac{\sqrt{3}}{2}b\sin(1.6^\circ)+1 \right)$.
From here, I think that to minimize $\theta$ I take two partial derivatives with respect to $a,b$. From this I have read that I can find the critical points of the differentiated $\arccos$ function, and then find the set of $a,b$ that overall minimize the problem. I just don't know how the initial constraint can be added into the problem.
It is likely that no rational pair $(a,b)$ can satisfy the equation (they belong to an ellipse of irrational coefficients).
If you lift this constraint, the system of equations is linear in $a,b$ and will have a solution for any $\theta$. Hence the smallest value is $-5°$.