4. 228 If the values $x=\sin \alpha, y=\sin \beta$ are given, then the expression
$$z=\sin (\alpha+\beta)$$

in general has four different values. Try to write the equation connecting $x, y$ and $z$ without using radicals or trigonometric functions. And find the values of $x$ and $y$ for which $z=\sin (\alpha+\beta)$ has fewer than four values.
[Solution] From the given information, we have
$$\cos \alpha= \pm \sqrt{1-x^{2}}, \cos \beta= \pm \sqrt{1-y^{2}}$$

Therefore, $\approx=\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$ has the following four values:
$$\begin{array}{l}
z_{1}=x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}, \\
-z_{1}=-x \sqrt{1-y^{2}}-y \sqrt{1-x^{2}}, \\
z_{2}=x \sqrt{1-y^{2}}-y \sqrt{1-x^{2}}, \\
-\approx_{2}=-x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}.
\end{array}$$

The equation
$$\left(z-z_{1}\right)\left(z+z_{1}\right)\left(z-z_{2}\right)\left(z+z_{2}\right)=0$$

has roots $\approx_{1}, -\approx_{1}, \approx_{2}, -\approx_{2}$. Removing the parentheses and simplifying, we get
$$z^{4}-\left(z_{1}^{2}+z_{2}^{2}\right) z^{2}+z_{1}^{2} z_{2}^{2}=0$$

where
$$\begin{aligned}
z_{1}^{2}+z_{2}^{2} & =2\left[\left(x \sqrt{1-y^{2}}\right)^{2}+\left(y \sqrt{1-x^{2}}\right)^{2}\right] \\
& =2\left(x^{2}-2 x^{2} y^{2}+y^{2}\right) \\
z_{1} z_{2} & =\left(x \sqrt{1-y^{2}}\right)^{2}-\left(y \sqrt{1-x^{2}}\right)^{2} \\
& =x^{2}-y^{2}
\end{aligned}$$

Substituting these expressions into equation (1), we obtain the required equation:
$$z^{4}-2\left(x^{2}-2 x^{2} y^{2}+y^{2}\right) z^{2}+\left(x^{2}-y^{2}\right)^{2}=0$$

Since $\Delta=\left(x^{2}-2 x^{2} y^{2}+y^{2}\right)-\left(x^{2}-y^{2}\right)^{2}=4 x^{2} y^{2}\left(1-x^{2}\right)\left(1-y^{2}\right)$, for $\sin (\alpha+\beta)$ to have fewer than four values, it is necessary and sufficient that $x=0$ or $y=0$ or $|x|=1$ or $|y|=1$.