Solve a trigonometric equation ($\sin(β/2)=β/4$)

Question

How can we solve the equation: $$\sin(\beta/2)=\beta/4$$ where $\beta $ is in radians?

Answer 1

As André Nicolas commented, there is no analytical solution and numerical methods (such as Newton) should be used.

I cannot resist the pleasure of showing you a $1400$ years old approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician and astronomer.

Applied to your case, this would lead to the equation $$\frac{8 \left(\pi -\frac{\beta }{2}\right) \beta }{5 \pi ^2-2 \left(\pi -\frac{\beta }{2}\right) \beta }=\frac \beta 4$$ which reduces to a quadratic the solution of it being given by $$\beta=-8+\pi +2 \sqrt{16+(4-\pi ) \pi }\approx 3.78954$$

If you look at this post where I was asking almost the same question, you will see that Christian Blatter proposed for the solution of $\sin(x)=a x$, the following approximation $$x\approx \sqrt{{\pi^2+5.95839 a - 15.828 a^2\over 1 + 2.60371 a + 0.690687 a^2}}$$ which, applied to your case ($x=\frac \beta 2$,$a=\frac 12$), would give $\beta \approx 3.79122$ while, for six significant figures, the solution is $\approx 3.79099$.

Similarly, after Christian Blatter, I made a small improvement using $$x\approx \sqrt{\frac{9.86775+4.91766 a-14.7794 a^2}{1+2.48744 a+0.633963 a^2}}$$ which, applied to your case , would give $\beta \approx 3.79117$.

Answer 2

Of course $\beta=0$ is a solution, but I guess you should call it trivial. Setting $\beta/2=x$, you can realize what is going on by looking at the graph of $y=\sin x - \frac{x}{2}$. You will see that two more solutions exist (the function is trivially odd in $x$). You will never find a closed form in terms of elementary functions, so all you can do is to prove that they exist and to estimate them by some numerical method.

Answer 3

For $\beta\neq 0$ we have $\text{sinc}\frac{\beta}{2}=\frac{1}{2}$ so $\beta\approx\pm 3.79009$ (according to Wolfram alpha).