5. (2003 Shandong Province Competition Question) Given that $\alpha, \beta$ are both acute angles, $\cos (\alpha+\beta)=-\frac{4}{5}$. If we set $\sin \beta=x, \cos \alpha=y$, then the functional relationship between $y$ and $x$ is ( ).
A. $y=-\frac{4}{5} \sqrt{1-x^{2}}+\frac{3}{5} x(0<x<1)$
B. $y=-\frac{4}{5} \sqrt{1-x^{2}}+\frac{3}{5} x\left(\frac{4}{5}<x<1\right)$
C. $y=-\frac{4}{5} \sqrt{1-x^{2}}-\frac{3}{5} x\left(\frac{4}{5}<x<1\right)$
D. $y=-\frac{4}{5} \sqrt{1-x^{2}}-\frac{3}{5} x(0<x<1)$
5. Choose B. Reason: Pay attention to exploring from the transformation of trigonometric function expressions and the form of angles, and make full use of the given information.

From the known conditions, we have $\sin (\alpha+\beta)=\frac{3}{5}, \cos \beta=\sqrt{1-x^{2}}$, so
$$
\begin{aligned}
y & =\cos \alpha=\cos [(\alpha+\beta)-\beta] \\
& =\cos (\alpha+\beta) \cdot \cos \beta+\sin (\alpha+\beta) \cdot \sin \beta \\
& =-\frac{4}{5} \sqrt{1-x^{2}}+\frac{3}{5} x .
\end{aligned}
$$
The range of $x$ should satisfy the system of inequalities
$$
\left\{\begin{array}{l}
0<x<1, \\
0<-\frac{4}{5} \sqrt{1-x^{2}}+\frac{3}{5} x<1
\end{array}\right.
$$

Solving this, we get $\frac{4}{5}<x<1$.