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

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). $\square$