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Using the Mean Value Theorem, prove that

$|\cos^2(b)-\cos^2(a)|\gt \frac{1}{4}|b-a| $

for all $a,b \in (\frac{\pi}{4},\frac{\pi}{3})$

So far, I have

Let $f(x)=\cos^2(x)$

Then $f(x)$ is continuous and differentiable on all $x\in \mathbb{R}$ so it is continuous and differentiable on $x\in (\frac{\pi}{4},\frac{\pi}{3})$.

Applying the Mean Value Theorem to $f(x)$, we have:

$\frac{\cos^2(b)-\cos^2(a)}{b-a}= -2\cos(x)\sin(x)$

$\frac{\cos^2(b)-\cos^2(a)}{b-a}= -\sin(2x)$

$|\cos^2(b)-\cos^2(a)| = -\sin(2x)|b-a|$

Now I do not know how to go from here, as on the interval $(\frac{\pi}{4},\frac{\pi}{3})$, $-\frac{\sqrt3}{2}\lt-\sin(2x)\lt -1$.

Or am I making a mistake?

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You have forgotten the absolute value signs and in the interval $$\frac{\pi}{4}<x<\frac{\pi}{3}$$ is $$|\sin(2x)|$$ greater than $$\frac{1}{4}$$ $$|\sin(2x)|$$ has the lowest value $$|\sin(\frac{2\pi}{3})|>\frac{1}{4}$$

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