# Can we solve for $\delta$ in $\frac{π}{2} - \frac{R}{r}β = \arcsin\biggl(\frac{(R+r)\sin(δ)}{r}\biggr)+\frac{R}{r}δ$?

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?

• You need a numerical method. Remember that $x=\cos(x)$ does not show analytical solutions. If you give test values for $r,R,\beta$, I could shos you. – Claude Leibovici Aug 5 '18 at 8:46
• You should ask for textbook recommendations in a separate question. That said, do a search first. I'm pretty sure that people have asked for such recommendations a few times. – Blue Aug 5 '18 at 9:39
• I have rolled-back the question, since your last edit changed the nature of the problem in a way that invalidated the efforts of an answerer. If you want to ask the question in your edit, post it separately. – Blue Aug 14 '18 at 12:03

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.