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Use the Mean Value Theorem to prove that for all $a,b\in(\frac\pi8,\frac\pi4)$ we have $$|\tan(2a)-\tan(2b)|\geq4|a-b|$$ If using a theorem, justify each of the hypothesis of the theorem before applying it.

This is what I think it is but I am really unsure and would like to see a detailed solution please.

Define function $f:[a,b]\rightarrow\mathbb{R}$ by $f(x)=\tan(2x)$ with $a,b\in(\frac\pi8,\frac\pi4)$.

Note that since, $\tan(x)$ and $2x$ are continuous functions on the real line, then the composed function $\tan(2x)$ is continuous by the algebra of continuous functions. Hence $f$ is continuous on interval $[a,b]$. Also since $\tan(x)$ and $2x$ are differentiable functions, by the algebra of differentiable functions, the composed function $\tan(2x)$ is differentiable. Hence $f$ is differentiable on $(a,b)$.

Applying the MVT, there exists $x\in(a,b)$ such that $$\frac{f(b)-f(a)}{b-a}=\frac{\tan(2b)-\tan(2a)}{b-a}=\frac2{\cos^2(2c)}=f'(c)$$

Now I would just say $0\leq\cos^2(2c)\leq1$ but looking at what $a$ and $b$ can be, this is $0<\cos^2(2c)<1$ but the question has a more or equal to. I don't understand how you can get that.

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If $c\in(\frac\pi8,\frac\pi4)$, then $2c\in(\frac\pi4,\frac\pi2)$ and:

$$ 0 =\cos^2\left(\frac{\pi}{2}\right) <\cos^2(2c) < \cos^2\left(\frac{\pi}{4}\right) = \frac{1}{2}. $$

Hence:

$$ \frac{2}{\cos^2(2c)} > 4.$$

Edit:

If $ a= b$, then

$$ |\tan (2a) - \tan (2b)| = 0 = 4|a-b|. $$

If $a\not= b$, proceed as you did with the application of MVT to obtain: $$ |\tan (2a) - \tan (2b)| > 4|a-b|. $$

So, no matter what the relationship between $a$ and $b$ is, as long as they are both in $(\frac\pi8,\frac\pi4)$ we have $$ |\tan (2a) - \tan (2b)| \geq 4|a-b|. $$

Edit2:

If $a> b$, we show $$ \frac{\tan (2a) - \tan (2b)}{a-b} = \frac{2}{\cos^2(2c)}, $$ for some $c\in (b,a)$.

If $a<b$, we show $$ \frac{\tan (2b) - \tan (2a)}{b-a} = \frac{2}{\cos^2(2c)}, $$ for some $c\in (a,b)$.

Concisely, using modulus, we say that for $a\not=b$ we have:

$$ \frac{|\tan (2b) - \tan (2a)|}{|b-a|} = \frac{2}{\cos^2(2c)}, $$ for some $c$ in $(a,b)$ or $(b,a)$.

Function $\tan$ is strictly increasing on $(\frac\pi4,\frac\pi2)$.

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  • $\begingroup$ OK so $$\frac{\tan(2b)-\tan(2a)}{b-a}=\frac2{\cos^2(2c)}>4$$ right? then what? $\endgroup$ – snowman Dec 3 '14 at 18:12
  • $\begingroup$ but the inequality in the question is not strict............................... $\endgroup$ – snowman Dec 3 '14 at 18:15
  • $\begingroup$ Yes. And it's strict inequality as we started out with $a\not= b$. It will be an equality if you start out with $a=b$. $\endgroup$ – ir7 Dec 3 '14 at 18:16
  • $\begingroup$ I seriously don't understand. can you write it down on your solution please. $\endgroup$ – snowman Dec 3 '14 at 18:17
  • $\begingroup$ and I don't see how the modulus signs come in. $\endgroup$ – snowman Dec 3 '14 at 18:19
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Since $a,b\in(\frac\pi8,\frac\pi4)$, we have $2c\in(\frac\pi4,\frac\pi2)$. Then $$ \cos\frac\pi2<\cos 2c<\cos\frac\pi4\implies\cos^22c\le\frac{1}{2} $$ and $$ \frac{1}{\cos^22c}\ge2. $$

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  • $\begingroup$ for your first inequality, why are the signs not strict? since c is in the open interval - not closed. $\endgroup$ – snowman Dec 3 '14 at 17:59
  • $\begingroup$ If $a<c<b$, then also $a\le c\le b$. $\endgroup$ – Julián Aguirre Dec 3 '14 at 21:35

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