According to Lagrange's Mean Value Theorem (LMVT), if a function $f(x)$ is continuous on $\left[a,b\right]$ and differentiable on $\left(a,b\right)$, then there exists some constant $c$ such that
$$f'(c) = \frac{f(b) – f(a)}{b-a} $$
Now, I have the following question:
Q) Using Lagrange's Mean value theorem, prove that $\sin(x) < x$ for $x > 0$.
This fact seems obvious from the graph of $\sin(x)$ and it's unit circle interpretation but I have no clue how I'd use LMVT to prove it. Any hints on how to start?
Best Answer
Let $x\in(0,2\pi].$ The derivative of $\sin(x)$ is $\cos(x)$, and so for some $c\in(0,x)$ we have $$\cos(c)=\frac{\sin(x)-\sin(0)}{x-0}=\frac{\sin(x)}{x}.$$ Since $\cos(c)<1$, the claim is proven for all $0<x\leq2\pi$. Then you can repeat the same argument, replacing $0$ by $2\pi$, and deduce the claim for all positive numbers.