# How do you find the critical numbers of y = sin^2 x?

We know that cos2x=cos^2x-sin^2x=>cos2x=1-2*sin^2x=> sin^2x=1/2[1-cos2x]

Hence $y$ becomes $y = \frac{1}{2} \cdot \left[1 - \cos 2 x\right]$

Now the critical values $c$ of $f$ are those if and only if

i. $f ' \left(c\right) = 0$
ii. $f ' \left(c\right)$ is undefined

So the critical values are those for which $\frac{\mathrm{dy}}{\mathrm{dc}} = 0$ hence

$\frac{\mathrm{dy}}{\mathrm{dc}} = \sin 2 c = 0 \implies 2 c = n \cdot \pi \implies c = n \cdot \frac{\pi}{2}$

where n is an integer.

The critical points are at $x = \frac{n \cdot \pi}{2}$