15. Given that the function $f(x)$ is a decreasing function defined on $(-\infty, 1]$, and for all real numbers $x$, the inequality $f(k-\sin x) \geqslant$ $f\left(k^{2}-\sin ^{2} x\right)$ always holds, then the value of $k$ is $\qquad$.
15. -1 From monotonicity, removing the function notation, we get
$$
\left\{\begin{array} { l } 
{ k ^ { 2 } - \operatorname { sin } ^ { 2 } x \leqslant 1 , } \\
{ k - \operatorname { sin } x \leqslant k ^ { 2 } - \operatorname { sin } ^ { 2 } x , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
k^{2} \leqslant 1+\sin ^{2} x, \text { (1) } \\
k^{2}-k+\frac{1}{4} \geqslant\left(\sin x-\frac{1}{2}\right)^{2},
\end{array}\right.\right.
$$

By the problem statement, (1) and (2) hold for all $x \in \mathbf{R}$, thus we have
$$
\left\{\begin{array}{l}
k^{2} \leqslant\left(1+\sin ^{2} x\right)_{\operatorname{max}}=1 . \\
k^{2}-k+\frac{1}{4} \geqslant\left(\sin x-\frac{1}{2}\right)_{\max }^{2}=\frac{9}{4},
\end{array} \Rightarrow k=-1 .\right.
$$