已知 $\sin \alpha+\sin \beta=\frac{4}{5}, \cos \alpha+\cos \beta=\frac{3}{5}$, 则 $\cos (\alpha-\beta)$ 的值为 ( D )
析] 由已知, 得 $(\sin \alpha+\sin \beta)^{2}+(\cos \alpha+\cos \beta)^{2}=\left(\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}=1$,

所以 $2+2(\cos \alpha \cos \beta+\sin \alpha \sin \beta)=1$,

即 $2+2 \cos (\alpha-\beta)=1$.

所以 $\cos (\alpha-\beta)=-\frac{1}{2}$.