Proving $|\sin x| \leqslant |x |$

Question

Is it possible to prove $|\sin x| \leqslant |x |$ with only trigonometric identities? (not with well-known calculus or geometry proofs)

Answer 1

If we have the following definitions

$$\cos(x) = \frac{z + z^{-1}}{2}$$ $$\sin(x) = \frac{z - z^{-1}}{2i}$$

where $z = e^{ix}$, then all finitary trig identities are reduced to rational polynomial identities with this definition. Let me show an example (of the strongest inequality possible that is close to what we want):

$$\begin{array}{rcl} \sin(x)^2 + \cos(x)^2 &=& \left(\frac{z + z^{-1}}{2}\right)^2 + \left(\frac{z - z^{-1}}{2i}\right)^2 \\ &=& \frac{z^2 + 2 + z^{-2}}{4} - \frac{z^2 - 2 + z^{-2}}{4} \\ &=& \tfrac{4}{4} \\ &=& 1 \end{array}$$

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The hidden reason that it is impossible to prove $|\sin(x)| < |x|$ with trig identities lies in the complex plane. All these trig identities hold uniformly for real and complex $x$. There are obstructions to this inequality in the complex plane:

$$|\sin(\sqrt{-1})| = 1.1752\ldots \not < 1 = |\sqrt{-1}|$$

This tells us that the inequality is something special about trigonometry on the real line, and because trig identities are unable to distinguish this we need a new tool (such as geometry, calculus for example).

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Lastly just let me note there are infinitary trig identities such as Euler's product $$ \frac{\sin(x)}{x} = \prod_{n = 1}^\infty\cos\left(\frac{x}{2^n}\right)$$ which do prove that $|\sin(x)| < |x|$ (to see this, just take the absolute value of both sides multiply up $|x|$ and note the infinite product is bounded by $1$ which is a consequence of the triangle identity applied to the Pythagorean trig identity)

Answer 2

The range of trigonometric identities should be specified. I will assume that by a trigonometric identity, we mean a formula that can be obtained within a finite times of algebraic operations form the following set of axioms:

We are given two functions $c : \Bbb{R} \to \Bbb{R}$ and $s : \Bbb{R} \to \Bbb{R}$ such that for $f(x) = c(x) + is(x)$,

1. $2\pi$ is the (least) period of $f(x)$.

2. For all $x \in \Bbb{R}$ we have $\left| f(x) \right| = 1$.

3. For all $x, y \in \Bbb{R}$ we have $f(x+y) = f(x)f(y)$.

4. $f(r \pi)$ lies in the first quadrant $\{(x, y) : x > 0, y > 0 \}$ for all $r \in \Bbb{Q} \cap \left( 0, \frac{1}{2} \right)$.

We immediately obtain

$$ c(x)^2 + s(x)^2 = 1, \quad c(x+y) = c(x)c(y) - s(x)s(y), \quad s(x+y) = s(x)c(y) + c(x)s(y). $$

More generally, 3 yields the multiple angle formula.

Now $f(0) = f(0)^2$ is non-zero, we have $f(0) = 1$ and in particular $c(0) = 1$ and $s(0) = 0$.

Also, $f(x)\overline{f(x)} = 1 = f(x)f(-x)$ implies $f(-x) = \overline{f(x)}$, or equivalently, $c(-x) = c(x)$ and $s(-x) = -s(x)$.

Since $1 = f(2\pi) = f(\pi)^2$ and $f(\pi) \neq 1$, we must have $f(\pi) = -1$ and thus $c(\pi) = -1$ and $s(\pi) = 0$.

Since $f\left(\frac{k\pi}{2n}\right)$ lies in the first quadrant for $n \geq 2$ and $0 < k < n$, it follows that $f\left(\frac{k\pi}{n}\right) = f\left(\frac{k\pi}{2n}\right)^2$ lies in the upper half-plane. By noting that

$$ f(\pi/n)^n = f(\pi) = -1, $$

$z = f(\pi/n)$ is a solution to the equation $z^n = -1$. But there is a unique solution $z$ to this equation such that $z, z^2, \cdots, z^{n-1}$ simultaneously lie in the upper half-plane, namely $z = e^{\pi i/n}$. Therefore

$$f(\pi / n) = e^{\pi i / n}$$

and it follows that $f(r\pi) = \exp(i r \pi)$ for any $r \in \Bbb{Q}$. This replicates any known special values for trigonometric functions which are attainable through some algebra on trigonometric identities.

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In this way we can obtain all the possible trigonometric identities for $c(x)$ and $s(x)$ instead of $\cos x$ and $\sin x$. Thus if we can prove that $|\sin x| \leq |x|$ from trigonometric identities, we should also have $|s(x)| \leq |x|$.

We prove that this is not the case. It is now clear that the axiom above is equivalent to the following assertion:

$f$ is a monomorphism from $G = \Bbb{R}/(2\pi\Bbb{Z})$ to $S^1$ such that $f(x) = \exp(ix)$ for all $x \in (2\pi\Bbb{Q})/(2\pi\Bbb{Z})$.

But assuming AC, we can find a function $f(x)$ satisfying the condition above yet $f(x) \neq \exp(i x)$ on $G$. It is not hard to prove that on any open set of $G$ the image $f(G)$ is dense in $S^1$. Thus we can find a sequence $x_n \in \Bbb{R}$ such that $x_n \to 0$ while $f(x_n) \to i$. That is, we have

$$ |s(x_n)| = 1 + o(1) > o(1) = |x_n|. $$

Therefore no such proof can exist and we have to exploit some analytic properties of trigonometric functions in order to prove the claim.

Answer 3

Clearly if $x >1$ then the inequality holds true.

Consider $0 \leq \theta < 1$

Consider the unit circle centered at $O$. Let $A$ be the point $(1,0)$ and B be the point ($cos(\theta), sin(\theta))$ Consider $\Delta OAB$ the area of this triangle is $\displaystyle\frac{sin(\theta)}{2}$ since the base is $1$, the radius, and height is $sin(\theta)$. While the area of sector $OB$ is just $\displaystyle\frac{\theta}{2}$. But the area of the sector is larger than the triangle. Hence

$\sin(\theta) \leq \theta$

I think this would help.

Answer 4

I think you cannot prove this with the given restrictions, since trigonometric identities that I can think of never mix angles and numbers, while the conclusion you want does want to compare angles and numbers. In other words you can think of angles having a separate dimension (in the physica sense of distance, weight, time, temperature etc.) and the trigonometric identities will hold true regardless of the unit of angle one chooses. However the relation $|\sin x|<|x|$ compares the number $\sin x$ with the angle $x$, and it will be false if you use a certain different unit of angle than radians (for instance if you would use $2\pi$ radians as unit, making the numeric value of $x$ smaller by a factor $2\pi$).

Answer 5

it's prove from that:

$cos(x)\leq \frac{sin(x)}{x} \leq 1$ then $|\frac{sin(x)}{x}|\leq 1 \Rightarrow |sin(x)|\leq |x|, \quad for \quad x\neq 0$

and as sin(0)=0 then

$|sin(x)|\leq |x|$.

the complete answer is in calulus silverman.