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?

 A: It s easier to deal with this form

$$ f(x) = \sin^2(x)\cos^2(x)=\frac{\sin^2(2x)}{4} $$

$$ \implies f'(x) = \sin(2x)\cos(2x)=\frac{\sin(4x)}{2}. $$
Now, you should be able to finish the problem.
A: Hint:
$$\begin{align}
0&=\cos x+\sin x\\
\sin x &= -\cos x \\
\text{Divide}&\text{ both sides by }\cos x\\
\tan x &= -1
\end{align}$$
Now you can solve for $x$ using inverse trigonometric functions.
A: 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
$\frac{1}{4} - x (1-x) = x^2 - x + \frac{1}{4} = (x-\frac{1}{2})^2 \ge 0$
A: if you use the Double angle formulae 
$$\sin^2(x)\cos^2(x)=(1-\cos2x)/2 \cdot (1+\cos2x)/2 = (1-\cos^2(2x))/4 = \sin^2(2x)/4 = (1-\cos4x)/8$$
so minimum is $1/8$ at $x=45\circ$ and maximum is 1/4 at $x=90\circ$
remember $\sin^2(x)=(1-\cos2x)/2$ and $\cos^2(x)=(1+\cos2x)/2$
A: 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 $\sin x=(e^{ix}-e^{-ix})/2i, \cos x=(e^{ix}-e^{-ix})/2$
