Is there a way to simplify $a\sqrt{1-a^2} + \arcsin(a) = \pi/4$?

Question

A while ago, I was eating pizza and wondered that if you were to cut parallel to one of the radii, how far along would you need to cut in order to split a slice's area in half?

In attempting to find a general answer for a sector with radius $r$ and angle $\theta$, I used some trigonometry to find $$a\sin\theta\sqrt{r^2 - a^2 \sin^2\theta} + r^2 \arcsin\left(\frac{a\sin\theta}{r}\right) - a^2\sin\theta\cos\theta = \frac{1}{2}r^2\theta$$ where $a$ is the distance along from the origin where we start our cut. After seeing this I had absolutely no idea how or even if I could continue, so I tried $r=1$ and $\theta = \pi/2$, i.e. a quarter-circle with radius 1. This gives: $$a\sqrt{1-a^2} + \arcsin(a) = \pi/4$$ Is there any way this expression can be further simplified?

Answer 1

If you let $x= \arcsin a$ then $a=\sin x$ and your equation is $$\sin x \cos x + x = \frac{\pi}{4}$$

or $$\sin 2x + 2x = \frac{\pi}{2}.$$

That seems like a transcendental equation that might have been studied.

Answer 2

Suppose that you want to solve for $a$ the equation $$a\sqrt{1-a^2} + \arcsin(a) =k$$ expand the lhs as a series to obtain $$k=2a-\sum_{n=1}^\infty \frac{|\beta_n|}{\gamma_n}\, a^{2n+1}$$ where the coefficients$\beta_n$ and $\gamma_n$ form respectively the sequences $A091154$ and $A143582$ in $OEIS$.

Using seriesreversion, the approximation $$a=\frac{1}{2}k+\frac{1}{48}k^3+\frac{13 }{3840}k^5+\frac{493 }{645120}k^7+\frac{37369}{185794560} k^9+O\left(k^{11}\right)$$

Tryin for a few values of $k \lt \frac \pi 2$, some results $$\left( \begin{array}{ccc} k & \text{estimate}&\text{solution} \\ 0.1 & 0.050021 & 0.050021 \\ 0.2 & 0.100168 & 0.100168 \\ 0.3 & 0.150571 & 0.150571 \\ 0.4 & 0.201369 & 0.201369 \\ 0.5 & 0.252716 & 0.252716 \\ 0.6 & 0.304787 & 0.304787 \\ 0.7 & 0.357786 & 0.357787 \\ 0.8 & 0.411963 & 0.411969 \\ 0.9 & 0.467630 & 0.467654 \\ 1.0 & 0.525184 & 0.525268 \\ 1.1 & 0.585145 & 0.585410 \\ 1.2 & 0.648200 & 0.648986 \\ 1.3 & 0.715269 & 0.717509 \\ 1.4 & 0.787586 & 0.793961 \\ 1.5 & 0.866810 & 0.886571 \\ \end{array} \right)$$

Answer 3

I prefer to put a second answer for the first equation.

Let $a=x\,r$ and simplify. You then need to find the zero of function $$f(x)=\theta ^2-2 x \sin (\theta ) \sqrt{1-x^2 \sin ^2(\theta )}+x^2 \sin (2\theta )-2 \sin ^{-1}(x \sin (\theta ))$$ Hoping that $x$ is not too large, te Taylor expansion is $$f(x)=\theta ^2-4 x \sin (\theta )+x^2 \sin (2 \theta )+\frac{2}{3} x^3 \sin^3(\theta )+\frac{1}{10} x^5 \sin ^5(\theta )+O\left(x^7\right)$$ Using series reversion, an estimate is $$\color{red}{x=\frac{\theta ^2 \csc (\theta )}{2949120}\,\, \sum_{n=0}^5c_n\,\theta^{2n}}$$ where $$\color{blue}{\left( \begin{array}{cc} n & c_n \\ 0 & 737280 \\ 1 & 92160 \cot (\theta ) \\ 2 & 23040 \cot ^2(\theta )+7680 \\ 3 & 7200 \cot (\theta ) \csc ^2(\theta )-2400 \cot (\theta ) \\ 4 & 2520 \cot ^2(\theta ) \csc ^2(\theta )+312 \\ 5 & 28 \cot (\theta )+945 \cot (\theta ) \csc ^4(\theta )-630 \cot (\theta ) \csc ^2(\theta ) \\ \end{array} \right)}$$ and then some results

$$\left( \begin{array}{ccc} \theta & \text{approximation} & \text{solution} \\ 0.1 & 0.025362 & 0.025362 \\ 0.2 & 0.051643 & 0.051643 \\ 0.3 & 0.079136 & 0.079136 \\ 0.4 & 0.108134 & 0.108134 \\ 0.5 & 0.138936 & 0.138937 \\ 0.6 & 0.171846 & 0.171850 \\ 0.7 & 0.207172 & 0.207186 \\ 0.8 & 0.245233 & 0.245268 \\ 0.9 & 0.286352 & 0.286434 \\ 1.0 & 0.330859 & 0.331035 \\ 1.1 & 0.379088 & 0.379442 \\ 1.2 & 0.431377 & 0.432048 \\ 1.3 & 0.488066 & 0.489281 \\ 1.4 & 0.549468 & 0.551612 \\ 1.5 & 0.615792 & 0.619572 \\ \end{array} \right)$$

And we could do much better, be sure ! It is just the problem of expanding $f(x)$ to higher orders.

For example, expanding $f(x)$ up to $O(x^{15})$, the results are $$\left( \begin{array}{ccc} \theta & \text{approximation} & \text{solution} \\ 0.1 & 0.025362 & 0.025362 \\ 0.2 & 0.051643 & 0.051643 \\ 0.3 & 0.079136 & 0.079136 \\ 0.4 & 0.108134 & 0.108134 \\ 0.5 & 0.138937 & 0.138937 \\ 0.6 & 0.171850 & 0.171850 \\ 0.7 & 0.207186 & 0.207186 \\ 0.8 & 0.245268 & 0.245268 \\ 0.9 & 0.286434 & 0.286434 \\ 1.0 & 0.331035 & 0.331035 \\ 1.1 & 0.379441 & 0.379442 \\ 1.2 & 0.432044 & 0.432048 \\ 1.3 & 0.489269 & 0.489281 \\ 1.4 & 0.551572 & 0.551612 \\ 1.5 & 0.619454 & 0.619572 \\ \end{array} \right)$$

Edit

Looking at the very first term of the series expansion, we have a first estimate $$x_0=\frac{1}{4} \theta ^2 \csc (\theta )$$ Use the first iteration of Halley method and obtain $$x_{1} = x_0 - \frac {2 \,f(x_0)\, f'(x_0)} {2 \,{[f'(x_0)]}^2 - f(x_0)\, f''(x_0)}$$ where $$f'(x)=-4 \sin (\theta ) \left(\sqrt{1-x^2 \sin ^2(\theta )}-x \cos (\theta)\right)$$ $$f''(x)=4 \sin (\theta ) \left(\cos (\theta )+\frac{x \sin ^2(\theta)}{\sqrt{1-x^2 \sin ^2(\theta )}}\right)$$ The formula is a bit too long to type here but $x_1$ is totally explicit.

We know that $x_1$ is an underestimate of the root and, since $$f(x_1)\times f''(x_1)~> ~0$$ by Darboux theorem, the exact solution will be reached without any overshoot.

This is simpler to use and quite good. Repeating the table $$\left( \begin{array}{ccc} \theta & x_1 & \text{solution} \\ 0.1 & 0.025362 & 0.025362 \\ 0.2 & 0.051643 & 0.051643 \\ 0.3 & 0.079136 & 0.079136 \\ 0.4 & 0.108134 & 0.108134 \\ 0.5 & 0.138937 & 0.138937 \\ 0.6 & 0.171849 & 0.171850 \\ 0.7 & 0.207184 & 0.207186 \\ 0.8 & 0.245265 & 0.245268 \\ 0.9 & 0.286426 & 0.286434 \\ 1.0 & 0.331020 & 0.331035 \\ 1.1 & 0.379415 & 0.379442 \\ 1.2 & 0.432005 & 0.432048 \\ 1.3 & 0.489217 & 0.489281 \\ 1.4 & 0.551523 & 0.551612 \\ 1.5 & 0.619462 & 0.619572 \\ \end{array} \right)$$

Answer 4

There are many $x$ and $\sin(x)$ transcendental equations on this site, but why not add another answer to one of them using the incomplete beta function $\text B_x(a,b)$:

$$x=a\sqrt{1-a^2}+\sin^{-1}(a)=\int_0^{a^2}\frac{\sqrt{1-t}}{\sqrt t}dt=\text B_{a^2}\left(\frac12,\frac32\right)$$

Therefore, we use inverse beta regularized $\text I^{-1}_s(a,b)$

$$a=\pm\sqrt{\text I^{-1}_\frac{2x}\pi\left(\frac12,\frac32\right)}$$

Shown here. With $a=\frac\pi2,\frac\pi4$ etc, you transform the circle problem into a quartile or median of a T distribution CDF, in terms of beta regularized $\text I_x(a,b)$, with $3$ degrees of freedom.