# How to prove $\left | \sin(x)-\sin(c) \right |\leq \left | x-c \right |$ for $c$ constant [closed]

• Do you know that $| \sin x | \leq | x |, \: \forall x \in \mathbf{R}$? – user305860 Jun 7 '16 at 15:14
• You can use the mean value inequality. – BrL Jun 7 '16 at 15:17
• @ErikJoensson Yes – Man Big Jun 7 '16 at 15:20
• @BrL Can you show me specifically？ – Man Big Jun 7 '16 at 15:25

Prove that $$\sin \phi - \sin \psi = 2 \cos \frac{\phi + \psi}{2} \sin \frac{\phi - \psi}{2}$$ by using $\sin (\phi + \psi) - \sin (\phi - \psi) = 2 \cos \phi \sin \psi$. Then with $|\sin \phi | \leq |\phi|$ and the formula above, you obtain the result. Note that this is the general formula, where both $\phi$ and $\psi$ are variables.
• The $+$ in the last $\sin$ should be a $-$. – BrL Jun 7 '16 at 15:33
@Erik Joensson's proof is original and only uses basic trigonometry. You can also use the mean value theorem : there exists some $t$ such that $\sin(x)-\sin(c) = (x-c)\sin'(t) = (x-c)\cos(t)$. But $\cos$ is bounded by $1$.
Let us first consider when $x\ge c$, we see that we can drop the absolute value bars on the right hand side of the inequality leaving us with $$\pm(\sin x - \sin c)\le x-c$$ Notice that when $x=c$ they are both $0$ and equal. Taking the directional derivative in the direction of $<1,0>$ (just a regular derivative) of both sides we have $$\frac{d}{dx}\pm(\sin x - \sin c)=\pm\cos x \le 1=\frac{d}{dx}(x-c)$$ thus, since they are both equal at $x=c$, and the value of $x-c$ is increasing at a greater rate than $\sin x - \sin c$, we have shown that $$|\sin x-\sin c|\le|x-c|\qquad\text{for}\qquad x\ge c$$ Now when $x\le c$, we can drop the absolute value bars, but we must multiply both sides by $-1$ giving $$\sin c -\sin x\le c-x$$ Again at $x=c$ they are both $0$ and equal. This time taking the directional derivative in the direction of $<-1,0>$ of both sides we have $$\frac{d}{dx}\pm(\sin c -\sin x)\cdot<-1,0>=\pm\cos x\le 1=\frac{d}{dx}(c-x)\cdot<-1,0>$$ Thus we have shown the equality also holds for $x\le c$ and we are done.