# Prove $\ \sin(x) < x \ \ \ \forall x \in(0, 2\pi)$

Problem : Prove $\sin(x) < x \ \ \ \forall x \in(0, 2\pi)$

Now I have a possible solution for this, using limits and the first derivatives of $\sin(x)$ and $x$, but I don't feel it's a very rigorous or succinct way to prove this. Can any of you find better ways to prove this? My proof is listed below.

Possible Proof

First we take the limits of both $\sin(x)$ and $x$ as $x \to 0^{+}$.
$$\lim_{x \to \ 0^{+}}\ \sin(x) = \sin(0) = 0$$ $$\lim_{x \to \ 0^{+}}\ x = 0$$

Next we take the derivatives of $\sin(x)$ and $x$. To see how they increase/descrease over the interval $(0, 2\pi)$

$$\frac{d}{dx} \ \sin(x)\ = \cos(x)$$ $$\frac{d}{dx} \ x\ = 1$$

$$\text{However} \ \ \cos(x) < 1, \ \forall x \in (0, 2\pi)$$ $$\implies \frac{d}{dx} \ \sin(x) < \frac{d}{dx} \ x\ ,\ \ \forall x \in (0, 2\pi)$$

This shows that the magnitude at which $\sin(x)$ is increasing is less than that of $x$ over the interval $(0, 2\pi)$, therefore if $\sin(x) \not> x$ as $x \to 0^{+}$, $\sin(x) < x, \ \forall x \in (0, 2\pi)$

$$Q.E.D$$

Would you say that this is a satisfactory proof? It doesn't seem particularly satisfactory to me, and it doesn't seem rigorous enough or all that succinct.

Are there better or more efficient/clearer ways to prove this, or problems of these sort of nature? Also if you have any comments about my proof-writing skills please leave them below.

• It's much simpler to say that $g(x)=\sin(x)-x$ is such that $g(0)=0$ and that $g'<0$ over $(0,2\pi)$, because you have only one function to study, instead of two. Commented May 31, 2016 at 0:20
• You want to be very careful using derivatives to solve this. One often uses the fact that $\sin x < x$ in a right neighbourhood of $0$ in conjunction with the squeeze theorem to evaluate the limit $$\lim_{x \to 0} \frac{\sin x}{x}$$ There are thus circular reasoning issues. I am hesitant, however, to say that this is entirely incorrect; there are some definitions of $\sin$ where this proof is fine. Commented May 31, 2016 at 0:44
• Perhaps it is worth mentioning that several proofs of this inequality can be found here and here Commented May 31, 2016 at 4:27

You can use an easy corollary of the Mean value theorem:

Let $f$ and $g$ be continuous functions on an interval $[a,b]$, differentiable on $(a,b)$. If $f(a)\le g(a)$ and $f'(x)<g'(x)$ on $(a,b)$, then $f(x)<g(x)$ for all $x \in (a,b)$.

Note that actually, it is enough to prove it for the interval $(0,\frac\pi2)$.

Here is the easy way. Travel along the unit circle $x$ units in the positive direction (counterclockwise). You are at angle $x$ radians. The coordinates of the point are $(\cos(x), \sin(x))$.

The distance from the point to the $x$-axis is $\sin(x)$ for $0 < x < \pi$; this is shorter than the distance along the circular arc, which is $x$. Therefore, in the first (and second quadrants) you have you have $\sin(x) < x$. No calculus is required.

• This is classic and fully rigorous proof for $\sin x < x$. But it does require calculus /analysis in the justification of existence of length of an arc. +1 Commented May 31, 2016 at 4:39

If you know that $\cos x < 1$ for $0 < x < 2\pi$, then $$0 < \int_0^{x} 1-\cos(t) \, dt = x -\sin(x)$$

This isn't a rigorous proof, but you could also think of it this way: Consider $y=\sin(x)$ and $y=x$ $$\sin(0)=0$$ Now, the rate of growth of $y=\sin(x)$ is $\cos(x)$ while the rate of growth of $y=x$ is $1$. In the interval $(0,2\pi), \cos(x)<1$ Thus in the interval $(0,2\pi), \sin(x)<x$