Find angle given ratio between sine and angle

Question

I know an angle is between 0 and $\pi$ (180 degrees). I know the ratio between its sine and the angle itself. Specifically it's $\frac{15}{16}$, but I am more interested in the general case. Since there is a one-to-one correspondence between $\theta$ and $\frac{\sin\theta}{\theta}$ in ($0$, $\pi$], I should be able to find the angle given that ratio, but I'm really struggling to. Any ideas?

Answer 1

To be an invertible function does not always mean that the inverse function is easy to compute. Your problem is an instance of Kepler's equation arising in orbital mechanics, which is usually solved numerically through Newton's method and variants (see Colwell).

In our case, assuming $\theta\in(0,\pi)$ and $\frac{\sin\theta}{\theta}=\frac{15}{16}$, a good starting point for Newton's method is given by approximating $\frac{\sin\theta}{\theta}$ with $1-\frac{\theta^2}{6}$. By setting $\theta_0=\sqrt{\frac{3}{8}}$ Newton's method $$\theta_{n+1} = \frac{\theta_n\cos\theta_n-\sin\theta_n}{\cos\theta_n-\frac{15}{16}} $$

converges pretty fast to the actual solution $\theta=0.618254969154963\ldots\approx 35^\circ 25'24''$.
As a curious coincidence, such solution is pretty close to the golden ratio $\frac{\sqrt{5}-1}{2}$.

Answer 2

I guess what you are asking is how to compute the inverse function of $f(\theta) = \frac{\sin \theta}{\theta}$.

There's no particular formula for it. You could use analytic methods to estimate it, for example by computing the first few terms of its Taylor series expansion near a value that you know (as suggested in the link provided by @JackD'Aurizio's comment).

It's not always easy to compute the function you want to compute.

Think about how hard it is to compute just the inverse function of $\sin \theta$. Having a name for it, the "arcsin", does not make it any easier to compute (and one could assign a name for the inverse function of $\frac{\sin\theta}{\theta}$, but that would not make it any easier to compute).

Actually, for that matter, think about how hard it is to compute $\sin(\theta)$ itself. Being one of the most venerable functions of mathematics does not make it particularly easy to compute, although generally we can estimate it pretty well by analytic methods.

Answer 3

As already answered, you need a numerical method for the solution of $\sin(x)=k x$.

You can generate a pretty good estimate of the solution using the approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ was proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (have a look here.

This would reduce to a quadratic $$4 k x^2-(4 \pi k-16) x+\pi(5 \pi k-16 )=0$$ the solution looked for for being $$x=\frac{1}{120} \left(60 \pi -256+\sqrt{65536+30720 \pi -14400 \pi ^2}\right)\approx 0.613716$$ from which Newton method would converge very fast $$\left( \begin{array}{cc} 0 & \color{red}{0.61}371595846525587708 \\ 1 & \color{red}{0.618}30451936663628509 \\ 2 & \color{red}{0.6182549}7495721438489 \\ 3 & \color{red}{0.618254969154963}32702 \\ 4 & \color{red}{0.61825496915496324745} \end{array} \right)$$

If you want a better approximation, have alook here where Christian Blatter proposed $$x\approx\sqrt{{\pi^2+5.95839 k - 15.828 k^2\over 1 + 2.60371 k + 0.690687 k^2}}$$ leading to $x\approx 0.617646$

If you are ready for iterations, you could make a very simple approximation $\sin(x)\sim a (\pi-x)x$ and compute $a$ minimizing with respect to $a$ $$\int_0^\pi \left(\sin(x)-a (\pi-x)x \right)^2\,dx$$ leading to $a=\frac{120}{\pi ^5}$ and then an estimate $x=\pi -k\frac{\pi ^5 }{120}$ which, for your case would be $\approx 0.750814$ Edit

Back to the problem three and half years latter, continuing from Christian Blatter's answer, I built the following approximation $$x=\sqrt{\frac{\pi ^2+58.834 k+73.7272 k^2-49.5852 k^3-79.9966 k^4-12.849 k^5 } {1+7.96113 k+20.3924 k^2+19.167 k^3+5.64183 k^4+0.287409 k^5}}$$ in which all parameters are highly significant from a statistical point of view.

The errors are smaller than $10^{-8}$ as shown in the table below $$\left( \begin{array}{ccc} k & \text{exact} & \text{approximation} \\ 0.00 & \pi & \pi \\ 0.05 & 2.99145643340 & 2.99145643328 \\ 0.10 & 2.85234189445 & 2.85234189458 \\ 0.15 & 2.72114256303 & 2.72114256287 \\ 0.20 & 2.59573907964 & 2.59573907964 \\ 0.25 & 2.47457678737 & 2.47457678754 \\ 0.30 & 2.35644114986 & 2.35644114992 \\ 0.35 & 2.24032553177 & 2.24032553163 \\ 0.40 & 2.12534519139 & 2.12534519122 \\ 0.45 & 2.01067452551 & 2.01067452551 \\ 0.50 & 1.89549426703 & 1.89549426722 \\ 0.55 & 1.77893894148 & 1.77893894169 \\ 0.60 & 1.66003481326 & 1.66003481330 \\ 0.65 & 1.53761467622 & 1.53761467601 \\ 0.70 & 1.41018537665 & 1.41018537635 \\ 0.75 & 1.27569810928 & 1.27569810916 \\ 0.80 & 1.13110258565 & 1.13110258592 \\ 0.85 & 0.97135215453 & 0.97135215504 \\ 0.90 & 0.78668307205 & 0.78668307214 \\ 0.95 & 0.55191097862 & 0.55191097740 \\ 1.00 & 0 & 0 \end{array} \right)$$ For $k=\frac{15}{16}$, the approximate model would give $x=0.618254968311$.