# Any hints on how to prove that the function $\lvert\alpha\;\sin(A)+\sin(A+B)\rvert - \lvert\sin(B)\rvert$ is negative over the half of the total area?

I have this inequality with $$0 and a real $$\lvert\alpha\rvert<1$$: $$f(A,B):=\bigl|\alpha\;\sin(A)+\sin(A+B)\bigr| - \bigl| \sin(B)\bigr| < 0$$

Numerically, I see that regardless of the value of $$\alpha$$, the area in which $$f(A,B)<0$$ is always half of the total area $$\pi^2$$.

I appreciate any hints and comments on how I can prove this.

• TeX note: please use \lvert \rvert for absolute value, not \mid \mid. Compare $\lvert\alpha\rvert < 1$ \lvert\alpha\rvert < 1 to $\mid\alpha\mid < 1$ \mid\alpha\mid < 1. I have edited accordingly. Aug 16, 2021 at 21:18
• is $\alpha$ real or do we allow complex ones too? Aug 16, 2021 at 23:22
• It looks to me like there is a sort of symmetry involved between $A$ and $\pi-A$. Let $g(A)$ be the measure of the set of points in the square lying along the vertical line at $A$ such that $f<0$. Testing numerically, it looks to me like we always have $g(A)+g(\pi-A)=\pi$. Possibly this can just be proved using trig identities, I don't know.
– user21349
Aug 17, 2021 at 1:22
• @Ben one has $g(\alpha, A)+g(-\alpha, \pi-A)=\pi$ in the notation above and if $I(\alpha)$ is the area in cause, $I(\alpha)+I(-\alpha)=\pi^2$, $I(0)=I(1)=I(-1)=\pi^2/2$ Aug 17, 2021 at 3:47
• @BenCrowell: I think your idea is the right one . I made the same computation. And btw I'm also a rock climber AND a fan of Homer's Greek! :) Aug 17, 2021 at 22:16

Let us assume $$\alpha\in[0,1)$$ (the case of $$\alpha\in(-1,0]$$ is similar). As $$\sin B>0$$ for $$B\in (0,\pi)$$, the inequality $$f(A,B)<0$$ amounts to $$\alpha\sin A<\sin B-\sin(A+B),\quad -[\sin B+\sin(A+B)]<\alpha\sin A.\quad (\star)$$ Notice that $$\sin A=2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)$$, $$\sin B-\sin(A+B)=-2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}+B\right)$$ and $$\sin B+\sin(A+B)=2\cos\left(\frac{A}{2}\right)\sin\left(\frac{A}{2}+B\right)$$. Substituting in $$(\star)$$ and cancelling the positive terms $$2\sin\left(\frac{A}{2}\right)$$ and $$2\cos\left(\frac{A}{2}\right)$$, we obtain the equivalent inequalities $$\alpha\cos\left(\frac{A}{2}\right)<-\cos\left(\frac{A}{2}+B\right), \quad -\sin\left(\frac{A}{2}+B\right)<\alpha\sin\left(\frac{A}{2}\right).\quad (\star\star)$$ In $$(\star\star)$$, the LHS of the first inequality and the RHS of the second are non-negative. Hence $$\frac{A}{2}+B$$ - which belongs to $$\left(0,\frac{3\pi}{2}\right)$$ - must be in the second or the third quadrant; otherwise, the first inequality in $$(\star\star)$$ does not hold. Let us analyze these cases separately:

• If $$\frac{\pi}{2}\leq\frac{A}{2}+B\leq\pi$$, then the second inequality in $$(\star\star)$$ holds automatically (its RHS is always non-negative); and the first one can be written as $$\alpha\cos\left(\frac{A}{2}\right)<\cos\left(\pi-\frac{A}{2}-B\right).$$ Applying the strictly decreasing function $$\cos^{-1}:[0,1]\rightarrow\left[0,\frac{\pi}{2}\right]$$ yields: $$\pi-\frac{A}{2}-B<\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right).$$ This of course implies $$\frac{A}{2}+B\geq\frac{\pi}{2}$$. But we also need $$\frac{A}{2}+B\leq\pi$$. Combining these, the bounds for $$B$$ in terms of $$A\in(0,\pi)$$ are given by $$\pi-\frac{A}{2}-\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right)\leq B\leq\pi-\frac{A}{2}.$$ The difference of the two bounds is $$\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right)$$. Consequently, the contribution to the area of $$\{(A,B)\mid f(A,B)<0\}$$ is $$\int_{\{(A,B)\mid f(A,B)<0,\, \frac{\pi}{2}\leq\frac{A}{2}+B\leq\pi\}}\mathbf{1}= \int_{0}^\pi\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right){\rm{d}}A.\quad (1)$$
• If $$\pi\leq\frac{A}{2}+B\leq\frac{3\pi}{2}$$, all terms appearing in $$(\star\star)$$ are non-negative. We first rewrite these inequalities as $$\alpha\cos\left(\frac{A}{2}\right)<\cos\left(\frac{A}{2}+B-\pi\right), \quad \sin\left(\frac{A}{2}+B-\pi\right)<\alpha\sin\left(\frac{A}{2}\right).$$ Next applying strictly monotonic functions $$\cos^{-1}:[0,1]\rightarrow\left[0,\frac{\pi}{2}\right]$$ and $$\sin^{-1}:[0,1]\rightarrow\left[0,\frac{\pi}{2}\right]$$ to them results in: $$\frac{A}{2}+B-\pi<\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\quad \frac{A}{2}+B-\pi<\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right).$$ Hence the upper bound $$B<\pi+\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}-\frac{A}{2}$$ which of course implies $$\frac{A}{2}+B\leq\frac{3\pi}{2}$$. But $$\pi\leq\frac{A}{2}+B$$ is also required. We therefore arrive at the bounds for $$B$$ in terms of $$A\in(0,\pi)$$: $$\pi-\frac{A}{2}\leq B\leq \pi+\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}-\frac{A}{2}.$$
The difference of the bounds is $$\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}$$. Therefore, the contribution to the area of $$\{(A,B)\mid f(A,B)<0\}$$ is $$\int_{\{(A,B)\mid f(A,B)<0,\, \pi\leq\frac{A}{2}+B\leq\frac{3\pi}{2}\}}\mathbf{1}\\ =\int_{0}^\pi\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}{\rm{d}}A.\quad (2)$$

Adding $$(1)$$ and $$(2)$$, the area of $$\{(A,B)\mid f(A,B)<0\}$$ turns out to be $$\int_{0}^\pi\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right){\rm{d}}A\\+ \int_{0}^\pi\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}{\rm{d}}A.\quad (\star\star\star)$$ So the question is if the quantity above coincides with $$\frac{\pi^2}{2}$$ for all $$\alpha\in [0,1)$$. First, we claim that the minimum above is $$\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)$$. Notice that: $$\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right) \leq\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right)\\ \Leftrightarrow \cos^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)+\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right)\geq\frac{\pi}{2};$$ and the cosine of the last angle appearing above is $$\left[\alpha\sin\left(\frac{A}{2}\right)\right]\left[\alpha\cos\left(\frac{A}{2}\right)\right] -\sqrt{1-\alpha^2\sin^2\left(\frac{A}{2}\right)} \sqrt{1-\alpha^2\cos^2\left(\frac{A}{2}\right)};$$ which is negative as $$\alpha\sin\left(\frac{A}{2}\right)<\sqrt{1-\alpha^2\cos^2\left(\frac{A}{2}\right)}$$ and $$\alpha\cos\left(\frac{A}{2}\right)<\sqrt{1-\alpha^2\sin^2\left(\frac{A}{2}\right)}$$ due to $$|\alpha|<1$$. We conclude that $$(\star\star\star)$$ is equal to $$\int_{0}^\pi\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right){\rm{d}}A+ \int_{0}^\pi\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right){\rm{d}}A.$$ Call the expression above $$h(\alpha)$$. The goal is to establish $$h(\alpha)=\frac{\pi^2}{2}$$ for any $$\alpha\in[0,1]$$. This is clear when $$\alpha=0$$, and so it suffices to show $$\frac{{\rm{d}}h}{{\rm{d}}\alpha}\equiv 0$$. One has $$\frac{{\rm{d}}h}{{\rm{d}}\alpha}= -\int_0^{\pi}\frac{\cos\left(\frac{A}{2}\right)}{\sqrt{1-\alpha^2\cos^2\left(\frac{A}{2}\right)}}{\rm{d}}A +\int_0^{\pi}\frac{\sin\left(\frac{A}{2}\right)}{\sqrt{1-\alpha^2\sin^2\left(\frac{A}{2}\right)}}{\rm{d}}A;$$ which is clearly zero because the change of variable $$A\mapsto\pi-A$$ indicates $$\int_0^{\pi}\frac{\cos\left(\frac{A}{2}\right)}{\sqrt{1-\alpha^2\cos^2\left(\frac{A}{2}\right)}}{\rm{d}}A =\int_0^{\pi}\frac{\sin\left(\frac{A}{2}\right)}{\sqrt{1-\alpha^2\sin^2\left(\frac{A}{2}\right)}}{\rm{d}}A.$$ This concludes the proof.

This is equivalent to \begin{align} |\alpha \sin A + \sin(A+B)|&<|\sin B|\\ ((\alpha+\cos B) \sin A + \cos A \sin B)^2&<(\sin B)^2\\ ((\alpha + \cos B)^2-\sin^2 B)\sin^2 A &<-2(\alpha+\cos\beta)\sin A \cos A \sin B\\ \frac{(\alpha + \cos B)^2-\sin^2 B}{\sin B} &<\frac{-2(\alpha+\cos\beta)\cos A}{\sin A}\\ \end{align}

So the area in question is the sum of the areas with $$\frac{\sin^2 B-(\alpha + \cos B)^2}{|\alpha + \cos B|\sin B} >2\cot A,\ \ \alpha+\cos B > 0$$ $$\frac{\sin^2 B-(\alpha + \cos B)^2}{|\alpha + \cos B|\sin B} >-2\cot A,\ \ \alpha+\cos B < 0$$

Since $$-\cot A=\cot(\pi-A)$$, the area in question equals the area with $$\frac{\sin^2 B-(\alpha + \cos B)^2}{|\alpha + \cos B|\sin B} >2\cot A$$ This is equivalent to $$1/\frac{|\alpha + \cos B|}{\sin B}-\frac{|\alpha + \cos B|}{\sin B} >1/\tan\left(\frac{A}{2}\right)-\tan\left(\frac{A}{2}\right)$$ and therefore to $$\frac{|\alpha + \cos B|}{\sin B} <\tan\left(\frac{A}{2}\right)$$ So the area in question can also be written as $${\cal A}(\alpha)=\int_{B=0}^{\pi} 2\arctan\frac{|\alpha + \cos B|}{\sin B} dB$$ Now it is easy to verify $${\cal A}(0)=\pi^2/2$$, and \begin{align} \frac{d{\cal A}(\alpha)}{d\alpha}&=\int_{0}^{\arccos(-\alpha)}\frac{2 \sin B\, dB}{1+2\alpha \cos B+\alpha^2}dB- \int_{\arccos(-\alpha)}^\pi\frac{2 \sin B\, dB}{1+2\alpha \cos B+\alpha^2}\\ &=\frac{1}{2\alpha}\log\frac{1+\alpha}{1-\alpha}-\frac{1}{2\alpha}\log\frac{1+\alpha}{1-\alpha}\\ &=0 \end{align} which leads to $${\cal A}(\alpha)=\pi^2/2$$ for all $$\alpha$$.

• On the first question, I am not saying that the two areas represent the same region, but that they have the same measure. I implicitly used the fact that $\{A \in [0,\pi]: f>-2\cot A\}$ is the same region as $\{A \in [0,\pi]: f>2\cot(\pi-A)\}$ and therefore has the same measure as $\{A \in [0,\pi]: f>2\cot A\}$. On the second question, it’s only after evaluating the integrals that I can see that $dA(\alpha)/d\alpha=0$. Did you get something other than $\pm\log((1+\alpha)/(1-\alpha))/(2\alpha)$ when you did the integrals?
– user44143
Aug 17, 2021 at 22:38

Let $$x,y$$ denote hereafter variables in the interval $$I:=[0,\pi]$$. Denote $$S:=\{-\sin(y)<\alpha\sin(x)+\sin(x+y)<\sin(y)\}\subset I^2$$ the set to be measured, and $$\Delta:=\{x+y<\pi\}\subset I^2$$.

Note that intersecting $$S$$ with $$\Delta$$ one inequality that defines $$S$$ is automatically satisfied, namely $$S\cap\Delta= \{-\sin(y)<\alpha\sin(x)+\sin(x+y)<\sin(y),\;x+y<\pi \}=$$$$=\{ \alpha\sin(x)+\sin(x+y)<\sin(y),\;x+y<\pi\}.$$

Similarly, the elementary inequality $$\sin(x+y)\le\sin(y)-\sin(x)$$ for $$x+y\ge\pi$$ gives $$S^c\cap\Delta^c= \{-\sin(y)\ge\alpha\sin(x)+\sin(x+y),\;x+y\ge\pi \}.$$

It is then straightforward to check that the area preserving affine transformation $$(x,y)\mapsto (\pi-x,x+y)$$ maps (a.e.) $$S\cap\Delta$$ onto $$S^c\cap\Delta^c$$, so they have the same area. Then $$|S|=|S\cap\Delta|+|S\cap\Delta^c|=|S^c\cap\Delta^c|+|S\cap\Delta^c|=|\Delta^c|=\frac12|I^2|,$$ ending the computation.

• Note that the $\sin$ function is no special: the same computation works for other functions with little assumptions Aug 17, 2021 at 22:46