# Any hints on how to prove that the function |αsin(A)+sin(A+B)|−|sin(B)|\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 $$0 and a real $$|α|<1\lvert\alpha\rvert<1$$:
$$f(A,B):=|αsin(A)+sin(A+B)|−|sin(B)|<0 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 $$α\alpha$$, the area in which $$f(A,B)<0f(A,B)<0$$ is always half of the total area $$π2\pi^2$$.

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

Let us assume $$α∈[0,1)\alpha\in[0,1)$$ (the case of $$α∈(−1,0]\alpha\in(-1,0]$$ is similar). As $$sinB>0\sin B>0$$ for $$B∈(0,π)B\in (0,\pi)$$, the inequality $$f(A,B)<0f(A,B)<0$$ amounts to
$$αsinA
Notice that
$$sinA=2sin(A2)cos(A2)\sin A=2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)$$,
$$sinB−sin(A+B)=−2sin(A2)cos(A2+B)\sin B-\sin(A+B)=-2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}+B\right)$$ and
$$sinB+sin(A+B)=2cos(A2)sin(A2+B)\sin B+\sin(A+B)=2\cos\left(\frac{A}{2}\right)\sin\left(\frac{A}{2}+B\right)$$.
Substituting in $$(⋆)(\star)$$ and cancelling the positive terms
$$2sin(A2)2\sin\left(\frac{A}{2}\right)$$ and $$2cos(A2)2\cos\left(\frac{A}{2}\right)$$, we obtain the equivalent inequalities
$$αcos(A2)<−cos(A2+B),−sin(A2+B)<αsin(A2).(⋆⋆) \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)$$, the LHS of the first inequality and the RHS of the second are non-negative. Hence $$A2+B\frac{A}{2}+B$$ - which belongs to $$(0,3π2)\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 $$π2≤A2+B≤π\frac{\pi}{2}\leq\frac{A}{2}+B\leq\pi$$, then the second inequality in $$(⋆⋆)(\star\star)$$ holds automatically (its RHS is always non-negative); and the first one can be written as
$$αcos(A2) Applying the strictly decreasing function $$cos−1:[0,1]→[0,π2]\cos^{-1}:[0,1]\rightarrow\left[0,\frac{\pi}{2}\right]$$ yields:
$$π−A2−B
This of course implies $$A2+B≥π2\frac{A}{2}+B\geq\frac{\pi}{2}$$. But we also need $$A2+B≤π\frac{A}{2}+B\leq\pi$$. Combining these, the bounds for $$BB$$ in terms of $$A∈(0,π)A\in(0,\pi)$$ are given by
$$π−A2−cos−1(αcos(A2))≤B≤π−A2. \pi-\frac{A}{2}-\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right)\leq B\leq\pi-\frac{A}{2}.$$
The difference of the two bounds is $$cos−1(αcos(A2))\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right)$$. Consequently, the contribution to the area of
$${(A,B)∣f(A,B)<0}\{(A,B)\mid f(A,B)<0\}$$ is
$$∫{(A,B)∣f(A,B)<0,π2≤A2+B≤π}1=∫π0cos−1(αcos(A2))dA.(1) \int_{\{(A,B)\mid f(A,B)<0,\, \frac{\pi}{2}\leq\frac{A}{2}+B\leq\pi\}}\mathbf{1}= \int_{0}^\pi\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right){\rm{d}}A.\quad (1)$$
• If $$π≤A2+B≤3π2\pi\leq\frac{A}{2}+B\leq\frac{3\pi}{2}$$, all terms appearing in $$(⋆⋆)(\star\star)$$ are non-negative. We first rewrite these inequalities as
$$αcos(A2)
Next applying strictly monotonic functions
$$cos−1:[0,1]→[0,π2]\cos^{-1}:[0,1]\rightarrow\left[0,\frac{\pi}{2}\right]$$
and $$sin−1:[0,1]→[0,π2]\sin^{-1}:[0,1]\rightarrow\left[0,\frac{\pi}{2}\right]$$ to them results in:
$$A2+B−π
Hence the upper bound
$$B<π+min B<\pi+\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}-\frac{A}{2}$$
which of course implies $$\frac{A}{2}+B\leq\frac{3\pi}{2}\frac{A}{2}+B\leq\frac{3\pi}{2}$$. But $$\pi\leq\frac{A}{2}+B\pi\leq\frac{A}{2}+B$$ is also required. We therefore arrive at the bounds for $$BB$$ in terms of $$A\in(0,\pi)A\in(0,\pi)$$:
$$\pi-\frac{A}{2}\leq B\leq \pi+\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}-\frac{A}{2}. \pi-\frac{A}{2}\leq B\leq \pi+\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}-\frac{A}{2}.$$
The difference of the bounds is $$\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}$$. Therefore, the contribution to the area of
$$\{(A,B)\mid f(A,B)<0\}\{(A,B)\mid f(A,B)<0\}$$ is
$$\int_{\{(A,B)\mid f(A,B)<0,\, \pi\leq\frac{A}{2}+B\leq\frac{3\pi}{2}\}}\mathbf{1}\\ =\int_{0}^\pi\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}{\rm{d}}A.\quad (2) \int_{\{(A,B)\mid f(A,B)<0,\, \pi\leq\frac{A}{2}+B\leq\frac{3\pi}{2}\}}\mathbf{1}\\ =\int_{0}^\pi\min\left\{\cos^{-1}\left(\alpha\cos\left(\frac{A}{2}\right)\right),\sin^{-1}\left(\alpha\sin\left(\frac{A}{2}\right)\right)\right\}{\rm{d}}A.\quad (2)$$

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