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 $0<A,B<\pi$ 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.
 A: 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.
A: 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 $\alpha$.
A: 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.
