Finding Lipschitz for trigonometric functions

Question

how would i find the Lipshitz constant for $$\sin(x)\times \cos(x)$$

or other trigonometric functions?

How would I get my $\operatorname{abs}{x_1 - x_2}$

Answer 1

If a function is differentiable on some convex set $C$, then by the mean value theorem you have $|f(x)-f(y) | \le \sup_{c \in C} |f'(c)| |x-y|$ (with $x,y \in C$).

In your case, $f'(x) = \cos^2 x - \sin^2 x = \cos (2x)$, so we have $|f'(x)| \le 1$ for all $x \in \mathbb{R}$. Hence $|f(x)-f(y) | \le |x-y|$.

Answer 2

Notice that $\sin x \cos x = \frac{1}{2}\sin(2x)$. Then, $$\left|\frac{1}{2}\sin(2x) -\frac{1}{2} \sin(2y)\right| = \cos(2c) |x - y| \leq |x-y|,$$ where $c$ is between $x$ and $y$. Using the mean value theorem usually works, if the derivative of the function is bounded.

Answer 3

When a function is everywhere differentiable and its derivative is bounded, then it is Lipschitz-continuous, and the Lipschitz constant is the maximum absolute value of its derivative (or anything bigger).

This is a corrolary of the mean value theorem: Suppose $|f'(x)|\le M$ for all $x$. Then for every distinct $w,x$, there is some number $u$ between $w$ and $x$ such that $|f(w)-f(x)|=|f'(u)||w-x|\le M|w-x|$.