How to prove that $|f(x) - f(y)| \leq \pi|x-y|$

Question

Question states

Prove that for all $x$ and $y\in (0, \frac{\pi}{2})$ the function $f(t) = \cos(t^2)$ satisfies the following inequality

$$|f(x) - f(y)| \le \pi |x-y|.$$

Use the mean value theorem

So derivative of $\cos(t^2) = -2\sin(t)\cos(t) = - \sin(2t)$

so $ -\sin(2t) = \dfrac {\cos(x^2) - \cos(y^2)}{x-y}.$

$\sin(2t)$ is bounded by $1$ and $-1$ so $|f(x) - f(y)| \le |x-y|.$

I'm missing $\pi$ there, could you correct me?

Answer 1

we have $$\frac{f(b)-f(a)}{b-a}=-\sin(\xi^2)2\xi$$

Answer 2

I thought it might be instructive to present a way forward that does not require use of calculus, but rather uses only standard trigonometry. To that end, we proceed.

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Using the Prosthaphaeresis Identity

$$\cos(a)-\cos(b)=-2\sin\left(\frac{a+b}2\right)\sin\left(\frac{a-b}2\right)$$

along with the inequalities $|\sin(\theta)|\le |\theta|$ and $|\sin(\theta)|\le 1$, we find with $a=x^2$ and $b=y^2$

$$\begin{align} |\cos(x^2)-\cos(y^2)|&=2\left|\sin\left(\frac{x^2+y^2}2\right)\sin\left(\frac{x^2-y^2}2\right)\right|\\\\ &\le |x^2-y^2|\\\\ &=|x+y|\,|x-y|\\\\ &\le \pi |x-y| \end{align}$$

since for $x,y\in (0,\pi/2)$, $0\le x+y\le \pi$.