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:
$$
\begin{aligned}
z_{1} & =x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}, \\
z_{1} & =-x \sqrt{1-y^{2}}-y \sqrt{1-x^{2}}, \\
z_{2} & =x \sqrt{1-y^{2}}-y \sqrt{1-x^{2}}, \\
-\approx_{2} & =-x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}} .
\end{aligned}
$$

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

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 desired equation:
$$
z^{4}-2\left(x^{2}-2 x^{2} y^{2}+y^{2}\right) z^{2}+\left(x^{2}-y^{2}\right)^{2}=0 \text {. }
$$

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 $|y|=1$.