## Calculus (3rd Edition)

$c=\frac{3\pi}{4}$
MVT states that $f'(c)=\frac{f(b)-f(a)}{b-a}$ Then, $-\sin c-\cos c=\frac{[\cos 2\pi -\sin 2\pi]-[\cos 0-\sin 0]}{2\pi -0}$ $\implies -\sin c-\cos c=0$ $\implies -\sin c= \cos c$ $\implies \tan c=-1$ We know that $\tan \frac{\pi}{4}=1$ and $-1=-\tan \frac{\pi}{4}=\tan (\pi-\frac{\pi}{4})=\tan \frac{3\pi}{4}$ $c=\frac{3\pi}{4}$ lies in the interval $(0,2\pi)$ and is one such point that satisfies the conclusion of the MVT. Note that we also could have chosen $\frac{7\pi}{4}$ as the point.