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<A,B<π and a real |α|<1:
f(A,B):=|αsin(A)+sin(A+B)||sin(B)|<0

Numerically, I see that regardless of the value of α, the area in which f(A,B)<0 is always half of the total area π2.

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

Answer

Let us assume α[0,1) (the case of α(1,0] is similar). As sinB>0 for B(0,π), the inequality f(A,B)<0 amounts to
αsinA<sinBsin(A+B),[sinB+sin(A+B)]<αsinA.()
Notice that
sinA=2sin(A2)cos(A2),
sinBsin(A+B)=2sin(A2)cos(A2+B) and
sinB+sin(A+B)=2cos(A2)sin(A2+B).
Substituting in () and cancelling the positive terms
2sin(A2) and 2cos(A2), we obtain the equivalent inequalities
αcos(A2)<cos(A2+B),sin(A2+B)<αsin(A2).()
In (), the LHS of the first inequality and the RHS of the second are non-negative. Hence A2+B - which belongs to (0,3π2) - must be in the second or the third quadrant; otherwise, the first inequality in () does not hold. Let us analyze these cases separately:

  • If π2A2+Bπ, then the second inequality in () holds automatically (its RHS is always non-negative); and the first one can be written as
    αcos(A2)<cos(πA2B). Applying the strictly decreasing function cos1:[0,1][0,π2] yields:
    πA2B<cos1(αcos(A2)).
    This of course implies A2+Bπ2. But we also need A2+Bπ. Combining these, the bounds for B in terms of A(0,π) are given by
    πA2cos1(αcos(A2))BπA2.
    The difference of the two bounds is cos1(αcos(A2)). Consequently, the contribution to the area of
    {(A,B)f(A,B)<0} is
    {(A,B)f(A,B)<0,π2A2+Bπ}1=π0cos1(αcos(A2))dA.(1)
  • If πA2+B3π2, all terms appearing in () are non-negative. We first rewrite these inequalities as
    αcos(A2)<cos(A2+Bπ),sin(A2+Bπ)<αsin(A2).
    Next applying strictly monotonic functions
    cos1:[0,1][0,π2]
    and sin1:[0,1][0,π2] to them results in:
    A2+Bπ<cos1(αcos(A2)),A2+Bπ<sin1(αsin(A2)).
    Hence the upper bound
    B<π+min
    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.

Attribution
Source : Link , Question Author : math2021 , Answer Author : KhashF

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