16. Let $f(x)$ be a function defined on $\mathbf{R}$, satisfying
$$
\left|f(x)+\cos ^{2} x\right| \leqslant \frac{3}{4},\left|f(x)-\sin ^{2} x\right| \leqslant \frac{1}{4} \text {. }
$$

Then the function $f(x)=$ . $\qquad$
16. $\sin ^{2} x-\frac{1}{4}$.

Notice,
$$
\begin{array}{l}
1=\sin ^{2} x+\cos ^{2} x \\
\leqslant\left|f(x)+\cos ^{2} x\right|+\left|f(x)-\sin ^{2} x\right| \\
\leqslant \frac{1}{4}+\frac{3}{4}=1 .
\end{array}
$$

Therefore, $\left|f(x)+\cos ^{2} x\right|=\frac{3}{4},\left|f(x)-\sin ^{2} x\right|=\frac{1}{4}$.
Thus, $f(x)=\sin ^{2} x-\frac{1}{4}$.