Finding minima and maxima of $\sin^2x \cos^2x$ I am trying to find the turning points of $\sin^2x \cos^2x$ in the range $0<x<\pi$.
So far I have:
$\begin{align} f'(x) & =\cos^2x \cdot 2\sin x \cdot \cos x + \sin^2x \cdot 2\cos x \cdot -\sin x \\
& = 2 \cos^3x \sin x - 2\sin^3x \cos x \\
& = 2 \sin x \cos x (\cos^2x - \sin^2x) \\
& = 2 \sin x \cos x (\cos x + \sin x) (\cos x - \sin x) 
\end{align}$
And, after lots of simplification, I think I've found that:
$f''(x) = 2 \left[\left( \cos^2x-\sin^2x \right)^2 - 4\sin^2x \cos^2 x\right]$
My questions are:


*

*How can I evaluate $0 = \cos x + \sin x$ and $0 = \cos x - \sin x$ without resorting to graph plotting?

*Are there trigonometric identities that I could have used to simplify either derivative?


One might broadly interpret "algebraically" to mean "without calculus." This is actually quite doable, and in my opinion less complicated than dealing with messy trigonometric equations. 
Note the following identities:
$f(x) = \sin^2 x \cos^2 x = \sin^2 x - \sin^4 x = \sin^2 x (1-\sin^2 x) \ge 0$
Equality holds when $\sin x = 0$ or $\pm 1,$ which occurs only for $x = \frac{\pi }{2}$ on $(0,\pi ).$ 
I claim that $f$ is maximized at $\frac{\pi }{6}.$ Since $0\le \sin^2 x \le 1$ for all real $x,$ it suffices to show that the function $g(x) = x (1-x)$ is maximized at $x=\frac{1}{2}$ on $[0,1].$ We can see this by noting that
$\frac{1}{4} - x (1-x) = x^2 - x + \frac{1}{4} = (x-\frac{1}{2})^2 \ge 0$
