7. The function $f(x)=\frac{\sin \left(x+45^{\circ}\right)}{\sin \left(x+60^{\circ}\right)}, x \in\left[0^{\circ}, 90^{\circ}\right]$, then the product of the maximum and minimum values of $f(x)$
Answer $\frac{2 \sqrt{3}}{3}$.
Since $f^{\prime}(x)=\frac{\sin 15^{\circ}}{\sin ^{2}\left(x+60^{\circ}\right)}>0$, $f(x)$ is strictly increasing, so the maximum value is $f\left(90^{\circ}\right)=\sqrt{2}$, and the minimum value is $f\left(0^{\circ}\right)=\frac{\sqrt{2}}{\sqrt{3}}$, their product is $\frac{2 \sqrt{3}}{3}$.
Note: Monotonicity can also be directly proven by definition.