<s>[INST] 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?

 [/INST] 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
$\frac{1}{4} - x (1-x) = x^2 - x + \frac{1}{4} = (x-\frac{1}{2})^2 \ge 0$
</s>