Answer
The tangent line is horizontal at the points $~~(\frac{\pi}{4},\sqrt{2})~~$ and $~~(\frac{5\pi}{4},-\sqrt{2})$
Work Step by Step
$y = sin~x+cos~x$
$y' = cos~x-sin~x$
If the tangent line is horizontal, then $y'=0$
We can find the values of $x$ where $y' = 0$:
$y' = cos~x-sin~x = 0$
$cos~x = sin~x$
$tan~x = 1$
$x = \frac{\pi}{4}, \frac{5\pi}{4}$
When $x = \frac{\pi}{4}$:
$y = sin~\frac{\pi}{4}+cos~\frac{\pi}{4}$
$y = \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}$
$y = \sqrt{2}$
When $x = \frac{5\pi}{4}$:
$y = sin~\frac{5\pi}{4}+cos~\frac{5\pi}{4}$
$y = -\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}$
$y = -\sqrt{2}$
The tangent line is horizontal at the points $~~(\frac{\pi}{4},\sqrt{2})~~$ and $~~(\frac{5\pi}{4},-\sqrt{2})$
