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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?

Martin Sleziak
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Nick
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    Other proofs of the same inequality can be found, for example, in these threads: http://math.stackexchange.com/questions/407517/the-sine-inequality-frac2-pi-x-le-sin-x-le-x-for-0x-frac-pi2, http://math.stackexchange.com/questions/125298/how-to-strictly-prove-sin-xx-for-0x-frac-pi2, http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 – Martin Sleziak Sep 21 '14 at 05:16

1 Answers1

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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.

Amitai Yuval
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