What relationship exists between $x$, $y$, and $z$ if

$$
\frac{x}{\sin t}=\frac{y}{\sin 2 t}=\frac{z}{\sin 3 t} ?
$$
1. Most often, we mean an equality relation, which we first write down based on two independent equations that can be derived from (1), by eliminating $t$. Of course, we assume that none of the denominators are 0, that is, $t$ is not an integer multiple of $\pi / 2$ or $\pi / 3$. On the one hand, we see that $x=y=z=0$ satisfies (1) for any (permitted) $t$, but it is not possible for $x, y$, and $z$ to be a mix of 0 and a value different from it. We also rule this out.

From the equality of the first two ratios in (1):

$$
\frac{y}{x}=\frac{\sin 2 t}{\sin t}=2 \cos t
$$

and from the first and the last ratio:

$$
\frac{z}{x}=\frac{\sin 3 t}{\sin t}=\frac{3 \sin t-4 \sin ^{3} t}{\sin t}=3-4 \sin ^{2} t=-1+4 \cos ^{2} t
$$

Substituting (2) into this, and rearranging:

$$
\frac{z}{x}=-1+\left(\frac{y}{x}\right)^{2}, \quad x^{2}-y^{2}+x z=0
$$

which is the desired relation involving all three variables.

2. We can also obtain inequalities between $x$ and $y$, as well as $x$ and $z$, based on (2) and (3):

$$
|\cos t|60^{\circ}$, we take the absolute value of the negative edge (or edges) and fit the cuboid so that this edge lies on the negative side of the corresponding axis.

As $t$ varies, the line of the body diagonal is expected to trace out a conical surface, whose apex is the origin. We sketch this conjecture.

We surround the origin with the parallelepiped bounded by the planes $x= \pm 1, y= \pm \sqrt{2}, z= \pm 1$, and find the common points of the surface given by equation (4) with the bounding 4 rectangles and 2 squares (Fig. 1, the front plane: $y=-\sqrt{2}$).

![](https://cdn.mathpix.com/cropped/2024_05_02_0d7c23c13f14d77bb43fg-1.jpg?height=647&width=783&top_left_y=2002&top_left_x=649)

Fig. 1

On the plane $x=1$, we get the points of the parabola $z=y^{2}-1$ for $|y| \leq \sqrt{2}$, excluding the points $(y=0, z=-1)$ and $(y= \pm 1, z=0)$, as these would correspond to $t=90^{\circ}$, $60^{\circ}$, and $120^{\circ}$. (On the plane $x=-1$, we get the mirror image of the previous arc about the origin, and similarly, the case $y=-\sqrt{2}, z=-1$ does not need to be examined separately.)

For $y=\sqrt{2}$, only the points $(x=1, z=1)$ and $(-1,-1)$ satisfy the resulting equation $x^{2}+x z=x(x+z)=2$ (since $|x| \leq 1,|z| \leq 1$, and thus $|x+z| \leq 2$).

For $z=1$, we get the two arcs of the hyperbola $x=y^{2}-x^{2}$, or equivalently,

$$
\frac{(x+0.5)^{2}}{0.25}-\frac{y^{2}}{0.25}=1
$$

the arc from the intersection point $(x=1 / 3, y=2 / 3)$ with the line $y=2 x$ - see (5) - to $(1, \sqrt{2})$, excluding the starting point, and its mirror image about the $y=0$ plane. (This is also indicated by (6).)

Another view of part of the cone is shown in Fig. 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_0d7c23c13f14d77bb43fg-2.jpg?height=507&width=469&top_left_y=722&top_left_x=817)

Fig. 2

According to all this, the transformation of the cone equation from the form (4) to the parametric form (1) excludes three isolated components, as well as a part of the surface between two other components.