Here is another (Ramanujan's type) trigonometric system of three equations.
Question: " If
$(\sin{x}\sin{y})^{1/4}+(\cos{x} \cos{y})^{1/4}=\sqrt{1+\sqrt{2}(\sin{2x} \sin {2y})^{1/4}} \tag1$
$(\sin{y}\sin{z})^{1/4}+(\cos{y} \cos{z})^{1/4}=\sqrt{1+\sqrt{2}(\sin{2y} \sin {2z})^{1/4}} \tag2$
then
$(\sin{x} \cos{z})^{1/4}+(\cos{x} \sin{z})^{1/4}=(8\sin{2y})^{1/12}\tag3$
One solution is:
$\sin{2y}=\sqrt{5}-2;$
$\sin{2x}=(\sqrt{5}-2)^3 (4+\sqrt{15})^2;$
$\sin{2z}=(\sqrt{5}-2)^3 (4-\sqrt{15})^2.$
To respond to Somos, I start with the values of $n=1,\cfrac{1}{2}, 2$.
For $n=1$ we have:
$\sin x=\sqrt{\frac{1}{2}+2 \cdot 12^{1/4}-108^{1/4}}$
$\sin y=\frac{\sqrt{2}}{2}$
$\sin z=\sqrt{\frac{1}{2}-2 \cdot 12^{1/4}+108^{1/4}}$.
For $n=\frac{1}{2}$ we have:
$\sin x=\sqrt{2} \cdot (\sqrt{2}-1)^{3/2} \cdot (2-\sqrt{3})\cdot(\sqrt{3}+\sqrt{2})^{2}$
$\sin y=\sqrt{2\sqrt{2}-2}$
$\sin z=\sqrt{2}\cdot (\sqrt{2}-1)^{3/2}\cdot (2+\sqrt{3})\cdot(\sqrt{3}-\sqrt{2})^{2}$.
For $n=2$ we have:
$\sin x= (\sqrt{2}-1)^{3}\cdot (2-\sqrt{3})^{2}$
$\sin y=(\sqrt{2}-1)$
$\sin z=(\sqrt{2}-1)^{3}\cdot (2+\sqrt{3})^{2}$.
