If $\sin\alpha\cos\beta=-0.5,$ then find the range of values of $\cos\alpha\sin\beta.$ 
If $\sin\alpha\cos\beta=-0.5,$ then find the range of values of $\cos\alpha\sin\beta.$

My Attempt:
$$-1\le\sin(\alpha+\beta)\le1\\-1\le\sin\alpha\cos\beta+\cos\alpha\sin\beta\le1\\-1\le-0.5+\cos\alpha\sin\beta\le1\\-0.5\le\cos\alpha\sin\beta\le1$$
The answer given is $[-0.5,0.5]$
Can you please confirm?
 A: You have proved correctly that $\cos\alpha\sin\beta\geqslant-\frac12$. And now you can do this:\begin{align}\sin(\alpha-\beta)\geqslant-1&\iff\sin\alpha\cos\beta-\cos\alpha\sin\beta\geqslant-1\\&\iff-\frac12-\cos\alpha\sin\beta\geqslant-1\\&\iff-\cos\alpha\sin\beta\geqslant-\frac12\\&\iff\cos\alpha\sin\beta\leqslant\frac12.\end{align}
A: To show that the range of $\cos \alpha \sin \beta$ is $[-1/2, 1/2]$, namely that
$$
S = \{ \cos \alpha \sin \beta \mid \alpha, \beta \in \mathbb{R}, \sin \alpha \cos \beta = -1/2 \} = [-1/2, 1/2],
$$
it is not only necessary to show that
$$
\cos \alpha \sin \beta = -1/2 \implies -1/2 \le \sin \alpha \cos \beta \le 1/2
$$
for all $\alpha, \beta \in \mathbb{R}$,
as shown in José Carlos Santos's answer,
which shows that $S \subseteq [-1/2, 1/2]$,
but also that, for all $z \in [-1/2, 1/2]$, there exists a pair $\alpha, \beta$ such that
$$
{\cos \alpha \sin \beta = -1/2} \quad \text{and} \quad
{\sin \alpha \cos \beta = z},
$$
which shows that $S \supseteq [-1/2, 1/2]$.
To show this, fix $z \in [-1/2, 1/2]$.  We would like to find $\alpha, \beta \in \mathbb{R}$ such that
$$
\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = z - 1/2
$$
and
$$
\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta = z + 1/2.
$$
We can always pick $x, y \in \mathbb{R}$ such that $\sin x = z - 1/2$ and $\sin y = z + 1/2$ since $\{z - 1/2, z + 1/2\} \subset [-1, 1]$.  Then,
$$
\alpha = \frac{x + y}{2} \quad \text{and} \quad
\beta  = \frac{x - y}{2}
$$
satisfy the constraints.
