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

Question

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.

Answer 1

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 $(0,\frac\pi2)$.

Answer 2

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.

Answer 3

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

Answer 4

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 $(0,2\pi), \sin(x)<x$