TITLE: 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 [11 upvotes]: 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.

REPLY [6 votes]: 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.