[Math] Finding minima and maxima of $\sin^2x \cos^2x$

calculustrigonometry

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:

  1. How can I evaluate $0 = \cos x + \sin x$ and $0 = \cos x – \sin x$ without resorting to graph plotting?
  2. Are there trigonometric identities that I could have used to simplify either derivative?

Best Answer

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.

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