Answer
$\vert sin~x-cos~x\vert \leq \sqrt{2}~~~$ for all $x$
Work Step by Step
$y = sin~x-cos~x$
Note that $y$ is continuous and differentiable for all values of $x$
We can find the points $x$ where the function is a maximum or a minimum:
$y' = cos~x+sin~x = 0$
$sin~x = -cos~x$
$tan~x = -1$
$x = -\frac{\pi}{4}+2\pi n, \frac{3\pi}{4}+2\pi n,~~~$ where $n$ is an integer
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 $x = \frac{3\pi}{4}$:
$y = sin~(\frac{3\pi}{4})-cos~(\frac{3\pi}{4})$
$y = \frac{\sqrt{2}}{2}-(-\frac{\sqrt{2}}{2})$
$y = \sqrt{2}$
The maximum value is $\sqrt{2}$ and the minimum value is $-\sqrt{2}$
Therefore, $~~\vert sin~x-cos~x\vert \leq \sqrt{2}~~~$ for all $x$
