Solve for $\delta$:
$$\frac{\pi}{2} - \frac{R}{r}β = \arcsin\left(\frac{(R+r)\sin \delta}{r}\right)+\frac{R}{r}δ$$
My problem is that I can't comprehend how to get both $\delta$s out of a trig operation at the same time. If I apply 'sine' to both sides, I understand the 'arcsine' will be canceled, but does this operation apply 'sine' to $\frac{R}{r}\delta$ on the right, making it $\sin\left(\frac{R}{r}\delta\right)$ ???
Is this problem a question of a trigonometric derivative?
My second question is perhaps a more important one. I do not have a textbook pertaining to trigonometry, much less trigonometric derivatives. The math book I currently own is about matrices and calculus, but only having to do with 'regular' derivatives; no trigonometry derivatives, obviously.
What is a good textbook for trigonometry (regular trig-algebra)? What is a good textbook for trigonometry derivatives? Is there a textbook that contains both of these subjects?
Thanks is advance.
Letting $Q = R/r$ and rearranging a bit gives
$$ \frac{\pi}{2} - Q\beta - Q\delta = \arcsin \bigl( (Q + 1) \sin \delta \bigr) $$
then taking $\sin(\cdot)$ of both sides and using $\sin (\pi/2 - x) = \cos x$ results in
$$ \cos \bigl( Q(\beta + \delta) \bigr) = (Q + 1) \sin \delta. $$
If $Q = 1$ then the equation can be solved exactly. Otherwise, you can rewrite the equation as
$$ \delta = f(\delta) = \arcsin \left( \frac{\cos \bigl( Q(\beta + \delta) \bigr)}{Q + 1} \right) $$
and repeatedly apply $f$ to some initial guess for a solution $\delta$. It turns out that this fixed-point iteration converges in this particular case.