How to prove that a point defined by trigonometric functions involving 4 parameters is inside a certain tetrahedron?

Question

This question has emerged in connection with an attempt to solve this recent question (see Remark below):

How can be proven (see result of simulation on 10000 points on graphics below) that a point with coordinates $$\begin{cases} x&=&\cos(\gamma-\delta)-\cos(\beta-\alpha)\\ y&=&\cos(\beta-\delta)-\cos(\alpha-\gamma)\\ z&=&\cos(\beta-\gamma)-\cos(\alpha-\delta) \end{cases}$$ where $\alpha,\beta,\gamma,\delta$ are arbitrary angles in $(0,2 \pi)$ is necessarily inside the regular tetrahedron defined by inequations : $$\begin{cases} \ \ x+y+z&>&-2\\ \ \ x-y-z&>&-2\\ -x+y-z&>&-2\\ -x-y+z&>&-2 \end{cases} \ \ ? \tag{1}$$

otherwise said, tetrahedron with vertices

$$\begin{pmatrix} 2\\ 2\\ 2 \end{pmatrix}, \begin{pmatrix} \ \ 2\\ -2\\ -2 \end{pmatrix},\begin{pmatrix} -2\\ \ \ 2\\ -2 \end{pmatrix},\begin{pmatrix} -2\\ -2\\ \ \ 2 \end{pmatrix}.$$

My attempts : I have used different trigonometric formulas without any success. I have also tried to use tetrahedral coordinates, with failure as well.

Remark : If the first formula in (1) is established, the upsaid question is solved by setting

$$A=(\cos(\alpha),\sin(\alpha)), B=(\cos(\beta),\sin(\beta)), C=(\cos(\gamma),\sin(\gamma)), D=(\cos(\delta),\sin(\delta)),$$

and replacing the inequality to be established ($AB^2+\cdots > -4$) by

$$(\cos(\alpha)-\cos(\beta))^2+(\sin(\alpha)-\sin(\beta))^2+\cdots > -4$$

$$2-2 \cos(\alpha-\beta)+\cdots > -4$$

Answer 1

Define $a$, $b$, $c$, $d$ as the complex exponentials of $\alpha$, $\beta$, $\gamma$, $\delta$, respectively. That is, for instance, $$a := \cos\alpha+i\sin\alpha \qquad\qquad \cos\alpha=\frac12\left(a+a^{-1}\right) \qquad \sin\alpha=\frac12\left(a-a^{-1}\right)\tag{1}$$ Then the condition $x+y+z+2$ becomes, with a little massaging, $$(a-b-c-d)(\overline{a}-\overline{b}-\overline{c}-\overline{d}) = |a-b-c-d|^2 \tag{2}$$ so that we're assured $x+y+z+2$ is non-negative, which is almost what we want. There are edge cases of equality, however.

Setting $(2)$ equal to $0$ implies that $a=b+c+d$. (For $x-y-z+2$, $-x+y-z+2$, $-x-y+z+2$, we can isolate $b$, $c$, $d$ in the same way.) In any case, we have that one point on the unit circle is the sum of three other points.

To see that this is a rare occurrence, fix $b$ and $c$ (and therefore also $b+c$). Varying $d$, the locus of $b+c+d$ is a circle about $b+c$ that passes through $b$ and $c$. For $b+c=0$, that circle is the unit circle, and only $a=d$ satisfies $(2)$; otherwise, $a$ can only coincide with either $b$ or $c$. We can avoid this edge case ---so that $(2)$ is strictly positive--- simply by requiring $a$ to be distinct from $b$, $c$, $d$; to cover the counterparts of $(2)$ corresponding to the other tetrahedral planes, we require all of $a$, $b$, $c$, $d$ to be distinct. This amounts to requiring the same of $\alpha$, $\beta$, $\gamma$, $\delta$ (a reasonable restriction, given the origin of the problem). $\square$

Answer 2

This answer is only reducing the inequalities to the one shown in the link in the OP.

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I will use $a,b,c,d$ for $\alpha,\beta,\gamma,\delta$, this saves typing. So we have to show for $$ \begin{aligned} x &= \cos(c-d)-\cos(\color{blue}a-b)\ ,\\ y &= \cos(d-b)-\cos(\color{blue}a-c)\ ,\\ z &= \cos(b-c)-\cos(\color{blue}a-d)\ , \end{aligned} $$ the inequality $x+y+z\ge -2$, and the other three inequalities obtained by replacing in all three cases the variable $\color{blue}a\to \color{blue}{a+\pi}$, together with only one of $b\to b+\pi$ ($x$ stays, $y,z$ change sign), then $c\to c+\pi$, then $d\to d+\pi$.

So let us show only $x+y+z\ge -2$. An $a$-shift in $b,c,d$ (and/or the circular symmetry of the linked post) lets us also assume w.l.o.g. $a=0$. This is $$ 2-\color{blue}{\cos b}-\color{blue}{\cos c} -\cos d \ge -\cos(c-d) -\cos(d-b)-\color{green}{\cos (b-c)}\ . $$ Let $S,X,Y$ be so that $b=X+\pi$, $-c=Y+\pi$, and $S=(d-b)+\pi$, so the above becomes equivalently: $$ 2+\color{blue}{\cos X}+\color{blue}{\cos Y} + \color{green}{\cos(X+Y)} \ge \cos S +\cos(X+S) +\cos(X+Y+S)\ . $$ The linked proof goes through this line. (All discussion is not more than an algebraic rephrase of the inequality. But at one point we have to stop reshaping the data, and attack the problem. My strategy would be again to consider the inequality as a minimizing problem for a function of three variables $X,Y,S$, then isolate one "weak variable" predisposed for attack, then minimize first w.r.t. to this variable. As it stays above, $S$ appears only three times, only on the R.H.S., so this may be the weak point. We expand to obtain on the R.H.S. an expression in $\sin S$, $\cos S$, that can be easily minimized / maximized. Only two variables remain...)

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Please do not accept this as an answer, there is already a better answer around. This is just a friendly remark, which had to be moved to the answer dialog box for space reasons, showing that the approach in the OP was very close to mine.