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?

Question

I have this inequality with $0<A,B<\pi$ 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.

Answer 1

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.

Answer 2

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 $\alpha$.

Answer 3

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.