Find the area of the region enclosed by $\frac{\sin x}{\sin y}=\frac{\sin x+\sin y}{\sin(x+y)}$ and the $x$-axis.

Question

Here is the graph of $\dfrac{\sin x}{\sin y}=\dfrac{\sin x+\sin y}{\sin(x+y)}$.

Find the area of the region enclosed by the curve and the $x$-axis, from $x=0$ to $x=\pi$.

Where the question came from, and why I think the answer is $\dfrac{\pi^2}{8}$

This question arose when I asked myself: "A triangle's vertices are three uniformly random points on a circle. The side lengths are, in random order, $a,b,c$. The triangle inequality tells us that $P(a+b<c)=0$. But what is $P\left(a+b<\left(\frac{a}{b}\right)c\right)$, given that $\frac{a}{b}>1$ ?"

(These kinds of questions often have rational probabilities, for example $P(ab<c^2)=\frac35$ as shown here; $P(\frac1a+\frac1b<\frac1c)=\frac15$ as shown here; for a unit circle $P(ab<c)=\frac12$ as shown here.)

I assumed that the circle is centred at the origin, and the vertices of the triangle are:
$A\space(\cos(-2y),\sin(-2y))$ where $0\le y<\pi$
$B\space(\cos(2x),\sin(2x))$ where $0\le x<\pi$
$C\space(1,0)$

I let:
$a=BC=2\sin x$
$b=AC=2\sin y$
$c=AB=|2\sin(x+y)|$

By symmetry, we only need to consider the region where $x+y<\pi$, so we can drop the modulus signs.

$P\left(a+b<\left(\frac{a}{b}\right)c\right)=P\left(\color{red}{\dfrac{\sin x}{\sin y}>\dfrac{\sin x+\sin y}{\sin(x+y)}}\right)$

Then I tried to express $y$ in terms of $x$, or $x$ in terms of $y$ (so that I could get an integral), but I failed. Wolfram is not helpful. I rotated the graph $45^\circ$ by letting $x\to x-y$ and $y\to x+y$, which gives $\frac{\tan x}{\tan y}=\frac{\cos x+\cos y}{\cos x-\cos y}$, but I still couldn't isolate $x$ or $y$. Wolfram is still not helpful.

A simulation of $10^7$ random triangles yielded a proportion of $0.49998$ satisfying $a+b<\frac{ac}{b}$ given that $\frac{a}{b}>1$, suggesting that the probability is $\frac12$. If so, then the area of the shaded region should be $\frac12\times\frac{\pi^2}{4}=\color{red}{\frac{\pi^2}{8}}\approx1.2337$.

(Now I am more interested in this area question, than the probability question.)

Update 1

Using the Wolfram link in @Masd's answer, I made a desmos graph of $y$ as a function of $x$. It looks just like the original curve, except it has gaps that I have been unable to bridge.

Here is the link to my desmos graph. If someone can bridge the gaps, then we can use desmos to integrate from $x=0$ to $x=\pi$, and see if the area could be $\frac{\pi^2}{8}$.

Update 2

@Masd's answer now reports a numerical integral of $1.23370055014$, which matches $\frac{\pi^2}{8}$ to $11$ decimal places. Is there a proof that the area is exactly $\frac{\pi^2}{8}$?

Answer 1

$${\sin x\over\sin y}=\frac{\sin x+\sin y}{\sin x \cos y+\sin y \cos x}\implies \sin^2 x\cos y+\sin x \cos x \sin y =\sin y \sin x+\sin^2y$$ If we substitute $g = \sin y$ and $v = \sin x$, we get: $$ v^2\sqrt{1-g^2}+vg\sqrt{1-v^2}=vg+g^2 $$ WolframAlpha gives us the solutions for $g$ (note that I replaced $x=g$ and $a=v$ ) : Here

EDIT

I have programmed a complex continuation to the function in a personal graphing calculator. I got this snapshot:

For some reason the negative part of function was restored whilst the positive wasn't, I also calculated the integral, which equaled $1.23370055014$, precisely $\pi^2\over 8 $

The integral

I will not lie the integral is massive: $$ I=\int_0^\pi\arcsin\left(\frac13\left(\sin (2x)-\sin x+f(x)-\frac{g(x)}{f(x)}\right)\right)dx $$ Where $$ f(x)=\left(\frac12\left(-h(x)+\sqrt{4g(x)^3+h(x)^2}\right)\right)^\frac13$$

$$g(x)=3\sin^2 x-(\sin 2x-\sin x)^2$$

$$h(x)=16\left(\cos x\right)\sin^5 x-10\left(\cos x\right)\sin^3x-24\sin^5 x-10\sin^3 x$$

Note that the gaps at @Dan's desmos link, are most likely to computational error and that the value of the $\arcsin$ is practically real on the interval $0$ to $-\pi$

As seen on the imaginary part of the graph (white):

It is strange seeing the rises in around $-0.5\pi$, but those are also most likely to computational error, as the desmos link @Dan offered lacks them

Answer 2

Changing coordinates

Let $R_\spadesuit$ be the region bounded by $y=0$ and $y=F(x)$ over the interval $x\in(0,\pi)$, where $y=F(x)$ satisfies

$$\frac{\sin x}{\sin y} = \frac{\sin x + \sin y}{\sin(x+y)} \tag{$\spadesuit$}$$

(i.e. the intimidating $\arcsin$ integrand in @Masd's post). We perform the following change of coordinates to rewrite and simplify $(\spadesuit)$ :

$$\begin{align*} (X,Y) = \left(\frac{x+y}{\sqrt2}, \frac{-x+y}{\sqrt2}\right) &\iff (x,y) = \left(\frac{X-Y}{\sqrt2}, \frac{X+Y}{\sqrt2}\right) \tag1 \\ (u,v) = \left(\cos\frac X{\sqrt2}, \cos \frac Y{\sqrt2}\right) &\iff (X,Y) = \left(\sqrt2\,\arccos u, \sqrt2\,\arccos v\right) \tag2 \\ (s,t) = \left(\frac uv, \sqrt{\frac{1-v^2}{1-u^2}}\right) &\iff (u,v) = \left(s \sqrt{\frac{t^2-1}{s^2t^2-1}}, \sqrt{\frac{t^2-1}{s^2t^2-1}}\right) \tag3 \end{align*}$$

$(1)$ rotates $R_\spadesuit$ counterclockwise about the origin by $\dfrac\pi4$ rad and, using the angle sum/difference and Pythagorean identities, transforms $(\spadesuit)$ to

$$\begin{align*} \frac{\sin \frac{X-Y}{\sqrt2}}{\sin \frac{X+Y}{\sqrt2}} &= \frac{\sin\frac{X-Y}{\sqrt2} + \sin\frac{X+Y}{\sqrt2}}{\sin\left(\sqrt2\, X\right)} = \frac{\cos\frac Y{\sqrt2}}{\cos\frac X{\sqrt2}} \\ \implies & \frac{\cos\frac X{\sqrt2}}{\cos\frac Y{\sqrt2}} + \frac{1 + \frac{\cos\frac X{\sqrt2}}{\cos\frac Y{\sqrt2}} \sqrt{\frac{1-\cos^2\frac Y{\sqrt2}}{1-\cos^2\frac X{\sqrt2}}}}{1 - \frac{\cos\frac X{\sqrt2}}{\cos\frac Y{\sqrt2}} \sqrt{\frac{1-\cos^2\frac Y{\sqrt2}}{1-\cos^2\frac X{\sqrt2}}}} = 0 \tag{$\heartsuit$} \end{align*}$$

Eliminate the trig expressions with $(2)$ to transform $(\heartsuit)$ to

$$\frac uv + \frac{1 + \frac uv \sqrt{\frac{1-v^2}{1-u^2}}}{1 - \frac uv \sqrt{\frac{1-v^2}{1-u^2}}} = 0 \tag{$\clubsuit$}$$

which motivates $(3)$ and reduces $(\clubsuit)$ to a trivial equation,

$$s + \frac{1+st}{1-st} = 0 \tag{$\diamondsuit$}$$

Let $R_\heartsuit$, $R_\clubsuit$, and $R_\diamondsuit$ be the resp. regions obtained by applying the transformations listed above. I've plotted them in red below (arranged top-to-bottom, left-to-right). Solve $(\diamondsuit)$ for $t$ as a function of $s$ to recover an equation for one of the boundaries of $R_\diamondsuit$,

$$t(s) = \dfrac{s+1}{s(s-1)}$$

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Determining boundaries of $R_\square$

The result of $(1)$ is straightforward, shifting the line $y=0$ to $Y=X$ and $y=F(x)$ to $Y=G(X)$. Admittedly, I'm eyeballing here and assuming that the rotated graph of $F$ is a single-valued function $G$, but I think this just boils down to showing $x > F(x)$ and $F'(x) < 1$. The vertices $(0,0)$ and $(0,\pi)$ in the $(x,y)$ plane correspond to $(0,0)$ and $\left(\dfrac\pi{\sqrt2},\dfrac\pi{\sqrt2}\right)$ in $(X,Y)$.

$(2)$ matches the lines $Y=X$ and $u=v$, while $G$ is transformed to another (presumably) single-valued function that permits us to write $v=H(u)$.

$(3)$ requires that $t\ge1$. Solving $t(s)=1$ we find $s=1\pm\sqrt2$, and we observe that $t(s)\stackrel{s\to1^+}\longrightarrow\infty$. Only the region to the right of the asymptote $s=1$ with the vertex at $\left(1+\sqrt2,1\right)$ has a finite area.

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Computing the area

In terms of double integrals, the area we want to find is given by

$$\begin{align*} \iint_{R_\spadesuit} dA &= \int_{x=0}^\pi \int_{y=0}^{F(x)} dy \, dx \\ \iint_{R_\heartsuit} dA &= \int_{X=0}^\tfrac\pi{\sqrt2} \int_{Y=X}^{G(X)} dY \, dX \\ \iint_{R_\clubsuit} dA &= \int_{u=0}^1 \int_{v=H(u)}^u \frac2{\sqrt{1-u^2}\sqrt{1-v^2}} \, dv \, du \\ \iint_{R_\diamondsuit} dA &= \int_{s=1}^{1+\sqrt2} \int_{t=1}^\tfrac{s+1}{s(s-1)} \frac2{s^2t^2-1} \, dt \, ds \end{align*}$$

We wrap up by bridging the gap to the integral(s) evaluated in the linked MO post.

$$\begin{align*} \int \frac 2{s^2t^2-1} \, dt &= -\frac2s \tanh^{-1}(st) + C \\ \implies \iint_{R_\diamondsuit} dA &= 2 \int_1^{1+\sqrt2} \left(\tanh^{-1}s - \tanh^{-1}\frac{s+1}{s-1}\right) \, \frac{ds}s \\ &= 2 \int_1^{1+\sqrt2} \tanh^{-1} \frac{1+2s-s^2}{1+s^2} \cdot \frac{ds}s \\ &= \boxed{\int_1^{1+\sqrt2} \log \frac{s+1}{s(s-1)} \cdot \frac{ds}s} \end{align*}$$

by the identity $\tanh^{-1}\alpha\pm\tanh^{-1}\beta=\tanh^{-1}\dfrac{\alpha\pm\beta}{1\pm\alpha\beta}$.

Answer 3

I really liked your intuition about the $\dfrac{\pi^2}{8}$ value of the area. I tried to achieve $y=f(x)$ but couldn't and so I decided to approximate the area as follows:

I determined the points $A, B$ and $C$ of the curve with which it was not difficult to calculate the area of the polygon $OABCF$ as the sum of the areas of the triangles $\triangle{OBF},\triangle{OBA}$ and $\triangle{BFC}$ (for the last two I calculated the points $D$ and $F$. The área it remains to have is very little and, of course, can be reduced of size if we iterate the same procedure. What I wanted to get was an approximation or an area greater than $\dfrac{\pi^2}{8}$ in which case your claim would have been rejected.

With our data, calculation gives for the area $A$ of the polygon $OABCF$ $$2A=0.64\times\pi+1.27263506159\times0.129495173655+2.13955616033\times0.0879317917479$$ $$A=1.18177720166$$ $$\frac{\pi^2}{8}-A=0.0519233484762\gt0$$ The area left to calculate is small. Note that the areas enclosed by the curve and the segments $OA$ and $CF$ are very small (they are almost not noticeable in the attached figure). I agree now with the OP's feeling. I hope it is true.

Answer 4

The answer is indeed $\dfrac{\pi^2}{8}$.

The proof is a combination of two answers at MO: @fedja's answer and @Iosif Pinelis' answer.