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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?
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    $\begingroup$ $\sin (2x) = 2\sin x\cos x$, so $f(x) = \frac14 \sin^2 (2x)$, then $f'(x) = \sin(2x)\cos (2x) = \frac12 \sin (4x)$, and $f''(x) = 2\cos (4x)$. $\endgroup$ – Daniel Fischer Aug 27 '13 at 20:32
  • $\begingroup$ @DanielFischer Aha, these are far easier to deal with, thanks! If your comment were an answer I'd accept it :) $\endgroup$ – hertzsprung Aug 27 '13 at 20:53
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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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  • $\begingroup$ Shouldn't the divisor of f(x) be 4 rather than 2? (See Daniel Fisher's comment on my question) $\endgroup$ – hertzsprung Aug 31 '13 at 12:04
  • $\begingroup$ @hertzsprung: It is done. Thanks for the comment. $\endgroup$ – Mhenni Benghorbal Aug 31 '13 at 21:16
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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

$\frac{1}{4} - x (1-x) = x^2 - x + \frac{1}{4} = (x-\frac{1}{2})^2 \ge 0$

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  • $\begingroup$ I've edited the question title, I think "algebraically" was misleading (I was trying to say that I wanted to do it without having to plot any functions) $\endgroup$ – hertzsprung Aug 27 '13 at 20:50
  • $\begingroup$ Understandable - I just found this to be a little less messy. $\endgroup$ – user17794 Aug 27 '13 at 20:52
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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.

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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 $\cos^2(x)=(1+\cos2x)/2$

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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 $\sin x=(e^{ix}-e^{-ix})/2i, \cos x=(e^{ix}-e^{-ix})/2$

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