16. Given that when $x \in[0,1]$, the inequality $x^{2} \cos \theta-x(1-x)+(1-x)^{2} \sin \theta>0$ always holds, find the range of values for $\theta$.
16. The original inequality is transformed into $(\sin \theta+\cos \theta+1) x^{2}-(2 \sin \theta+1) x+\sin \theta>0$.

Let $f(x)=(\sin \theta+\cos \theta+1) x^{2}-(2 \sin \theta+1) x+\sin \theta, x \in[0,1]$. To ensure that $f(x)>0$ for all $x \in[0,1]$, it is necessary that $f(0)>0$ and $f(1)>0$, i.e., $\sin \theta>0$ and $\cos \theta>0$, which means $\theta$ is an angle in the first quadrant, and $\sin \theta+\cos \theta+1>0$. In this case, the parabola opens upwards, and its axis of symmetry is given by $x=\frac{2 \sin \theta+1}{2(\sin \theta+\cos \theta+1)}$.
Since $2(\sin \theta+\cos \theta+1)>2 \sin \theta+1>0$, it follows that $00$ to always hold, the minimum value of $f(x)$ must be positive.
In summary, $f(x)>0$,
$$
\begin{array}{l}
\Leftrightarrow\left\{\begin{array} { l } 
{ \sin \theta > 0 , } \\
{ \cos \theta > 0 , } \\
{ 4 ( \sin \theta + \cos \theta + 1 ) \sin \theta - ( 2 \sin \theta + 1 ) ^ { 2 } > 0 . }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
2 k \pi<\theta<2 k \pi+\frac{\pi}{2}, \\
k \pi+\frac{\pi}{12}<\theta<k \pi+\frac{5}{12} \pi .
\end{array}\right.\right. \\
\Leftrightarrow\left\{\begin{array}{l}
2 k \pi+\frac{\pi}{12}<\theta<2 k \pi+\frac{5}{12} \pi(k \in \mathbf{Z}) .
\end{array}\right.
\end{array}
$$

Therefore, $2 k \pi+\frac{\pi}{12}<\theta<2 k \pi+\frac{5}{12} \pi(k \in \mathbf{Z})$.