Answer
$$
f(t)=2\cos t + \sin 2t, \text { on } \,\, [0, \frac{\pi}{2}]
$$
The absolute maximum value is
$$
f\left(\frac{\pi}{6}\right)=\frac{3}{2}\sqrt{3} \approx 2.60,
$$
and the absolute minimum value is
$$
f(\frac{\pi}{2})=0.
$$
Work Step by Step
$$
f(t)=2\cos t + \sin 2t, \text { on } \,\, [0, \frac{\pi}{2}]
$$
Since $ f$ is continuous on $ [0, \frac{\pi}{2}], $ we can use the Closed Interval Method:
$$
f(t)=2\cos t + \sin 2t,
$$
$$
\begin{aligned}
f^{\prime}(t) &=-2 \sin t+\cos 2 t \cdot 2 \\
&=-2 \sin t+2\left(1-2 \sin ^{2} t\right) \\
&=-2\left(2 \sin ^{2} t+\sin t-1\right) \\
&=-2(2 \sin t-1)(\sin t+1)
\end{aligned}
$$
Since $f^{\prime}(t)$ exists for all $t$, the only critical numbers of $f$ occur when $f^{\prime}(t)=0,$ that is,
$$
\begin{aligned}
f^{\prime}(t) &=-2(2 \sin t-1)(\sin t+1)=0
\end{aligned}
$$
$ \Rightarrow $
$$
\sin t= \frac{1}{2} \quad\text {or } \quad \sin t=-1
$$
$ \Rightarrow $
$$
t=\frac{\pi}{6}.
$$
Notice that this critical number in the interval $[0, \frac{\pi}{2}], $
The values of $f $ at these critical numbers are
$$
f\left(\frac{\pi}{6}\right)=\sqrt{3}+\frac{1}{2}\sqrt{3}=\frac{3}{2}\sqrt{3} \approx 2.60,
$$
The values of $f$ at the endpoints of the interval are
$$
f(0)=2 ,\quad f(\frac{\pi}{2})=0.
$$
Comparing these four numbers, we see that the absolute maximum value is
$$
f\left(\frac{\pi}{6}\right)=\frac{3}{2}\sqrt{3} \approx 2.60,
$$
and the absolute minimum value is
$$
f(\frac{\pi}{2})=0.
$$
