<s>[INST] 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.
 [/INST] 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 $\alpha$.
</s>