8. The real value range of the function $f(x)=\sqrt{\cos ^{2} x-\frac{3}{4}}+\sin x$ is
This problem's difficulty: slightly difficult for the first attempt
Mainly tests: substitution method, function monotonicity
Solution: $\left[-\frac{1}{2}, \frac{\sqrt{2}}{2}\right]$
Let $\sin x=\frac{1}{2} \sin \varphi$, under this transformation, both variables simultaneously make $\mathrm{f}$ take real or complex values at corresponding values, so $\mathrm{f}$, as a real function, has its natural domain changed but its range remains unchanged under the transformation. Thus, $\cos ^{2} x=1-\frac{1}{4} \sin ^{2} \varphi, \sqrt{\cos ^{2} x-\frac{3}{4}}+\sin x=\frac{|\cos \varphi|+\sin \varphi}{2}$. It is easy to see that $h(\varphi)=\frac{|\cos \varphi|+\sin \varphi}{2}$ is a function with a period of $2 \pi$ and a symmetry axis at $\frac{\pi}{2}$, so it is only necessary to discuss its values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Within this interval, the absolute value symbol can be directly removed, and by analyzing its derivative, it can be seen that $h$ is increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{4}\right]$ and decreasing on $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$. From this, the range of $\mathrm{f}$ can be obtained.