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?


The equation $\cos x+\sin x=0$ is really the same as $\cos x-\sin x=0$ once you make the transformation $x\mapsto x+\frac{\pi}2$, because $\cos (x+\frac{\pi}2)=-\sin x$ and $\sin (x+\frac{\pi}2)=\cos(x)$. The latter can be solved by resorting to $\cos ^2+\sin ^2=1$.
Edit: We can derive $\cos (x+\frac{\pi}2)=-\sin x$ and similar results (cf. Daniel Fischer's comment) from the well known formulas $\cos (x+y)=\cos x\cos y-\sin x\sin y$ and $\sin (x+y)=\cos x\sin y+\cos y\sin x$. These can either be proved geometrically, or by resorting to some definition of $\sin$ and $\cos$, for example $\sin x=(e^{ix}-e^{-ix})/2i, \cos x=(e^{ix}-e^{-ix})/2$
