Answer
The maximum is $f(0)=f(2\pi) =1$ and the minimum is $f( 5\pi/4) =-\sqrt{2}$
Work Step by Step
Given $$y=\cos \theta+\sin \theta, \quad[0,2 \pi]$$
Since
\begin{align*}
\frac{dy}{d\theta}&=\cos \theta-\sin \theta
\end{align*}
To find critical points
\begin{align*}
f'(\theta)&=0\\
\cos \theta-\sin \theta&=0
\end{align*}
Then $\theta= \pi/4,\ \ 5\pi/4$. Because $\pi/4,\ 5\pi/4\in [0,2\pi]$, then $f(\theta)$ has a critical point $\theta=\pi/4,\ \ 5\pi/4$, since
\begin{align*}
f(0)&=1 \\
f(\pi/4 )&=\sqrt{2}\\
f(5\pi/4)&=-\sqrt{2}\\
f(2\pi/)&=1
\end{align*}
Then the maximum is $f(0)=f(2\pi) =1$ and the minimum is $f( 5\pi/ ) =-\sqrt{2}$
