0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Claude Leibovici Aug 5 '18 at 8:46
  • $\begingroup$ 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. $\endgroup$ – Blue Aug 5 '18 at 9:39
  • $\begingroup$ 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. $\endgroup$ – Blue Aug 14 '18 at 12:03
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.