Absolute extrema via the Closed Interval Method

Since $f$ is continuous on the closed interval $[0, 2\pi]$, the Closed Interval Method applies: differentiate to get $f'(x) = \cos(x) - \sin(x)$, which is defined everywhere, so critical numbers occur only where $f'(x) = 0$. Setting $\cos(x) = \sin(x)$ gives $\tan(x) = 1$, so on $[0, 2\pi]$ the critical numbers are $x = \pi/4$ and $x = 5\pi/4$.

Now evaluate $f$ at the critical numbers and the endpoints: $f(0) = 1$, $f(\pi/4) = \sin(\pi/4)+\cos(\pi/4) = \sqrt{2}$, $f(5\pi/4) = \sin(5\pi/4)+\cos(5\pi/4) = -\sqrt{2}$, and $f(2\pi) = 1$. Comparing all four values, the largest is $\sqrt{2}$ at $x=\pi/4$ and the smallest is $-\sqrt{2}$ at $x=5\pi/4$.