# Is f(x)= cos(x+(5pi)/6)  increasing or decreasing at x=pi/4 ?

Dec 5, 2015

Increasing.

#### Explanation:

Find the first derivative. If $f ' \left(\frac{\pi}{4}\right) < 0$, then the function is decreasing at that point. If $f ' \left(\frac{\pi}{4}\right) > 0$, then the function is increasing at that point.

To find $f ' \left(x\right)$, we will need to use the chain rule:

$u = x + \frac{5 \pi}{6}$
$f \left(x\right) = \cos \left(u\right)$
$\frac{d}{\mathrm{dx}} \left[\cos u\right] = - \sin \left(u\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$f ' \left(x\right) = - \sin \left(x + \frac{5 \pi}{6}\right) \cdot \frac{d}{\mathrm{dx}} \left[x + \frac{5 \pi}{6}\right]$

Note that $\frac{d}{\mathrm{dx}} \left[x + \frac{5 \pi}{6}\right] = 1$, so:

$f ' \left(x\right) = - \sin \left(x + \frac{5 \pi}{6}\right)$

$f ' \left(\frac{\pi}{4}\right) = - \sin \left(\frac{\pi}{4} + \frac{5 \pi}{6}\right)$

$f ' \left(\frac{\pi}{4}\right) = - \sin \left(\frac{13 \pi}{12}\right)$

You could calculate $\sin \left(\frac{13 \pi}{12}\right)$, but I think a more elegant approach would be to recognize that $\frac{13 \pi}{12}$ is just barely larger than $\pi$, so a reference angle of $\frac{13 \pi}{12}$ would land in the third quadrant.

Thus, the sine of that angle would be NEGATIVE, and since there is a negative sign outside of the sine, $f ' \left(\frac{\pi}{4}\right)$ is POSITIVE, meaning the function is increasing when $x = \frac{\pi}{4}$.