How to use mean value theorem to prove the inequality $|\sin{x}-\sin{y}|\le|x-y|$ for all $x,y\in\Bbb{R}$?
So let us set $f(x)=\sin{x}$ then it's differentiable on $(x,y)$ and continuous on $[x,y]$. So there exists $c$ on $(x,y)$ such that $f'(c)=\frac{\sin{x}-\sin{y}}{x-y}$.
Since $\max{(\sin{x}-\sin{y})}$ is $2$ where $|x-y|=\pi\ge2$. So I guess then we make the conclusion.
Could someone check if my thinking is right?
Best Answer
1) $$\frac{\sin x-\sin y}{x-y}=cos(c)$$ where $c \in [y,x]$. ($y < x$)
2) Take absolute value.
3) Use $|\cos c|\le 1$.
4) Multiply with positive $|x-y|$.