3. The room temperature $T$ (unit: Celsius) as a function of time $t$ (unit: hours) is given by: $T=a \sin t+b \cos t, t \in(0,+\infty)$, where $a, b$ are positive real numbers. If the maximum temperature difference in the room is 10 degrees Celsius, then the maximum value of $a+b$ is $\qquad$ .
Answer: $5 \sqrt{2}$. Solution: By the auxiliary angle formula: $T=a \sin t+b \cos t=\sqrt{a^{2}+b^{2}} \sin (t+\varphi)$, where $\varphi$ satisfies the conditions $\sin \varphi=\frac{b}{\sqrt{a^{2}+b^{2}}}, \cos \varphi=\frac{a}{\sqrt{a^{2}+b^{2}}}$, then the range of the function $T$ is $\left[-\sqrt{a^{2}+b^{2}}, \sqrt{a^{2}+b^{2}}\right]$, the maximum indoor temperature difference is $2 \sqrt{a^{2}+b^{2}} \leq 10$, thus $\sqrt{a^{2}+b^{2}} \leq 5$. Therefore, $a+b \leq \sqrt{2\left(a^{2}+b^{2}\right)} \leq 5 \sqrt{2}$, equality holds if and only if $a=b=\frac{5}{2} \sqrt{2}$.