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

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



The hidden reason that it is impossible to prove $|\sin(x)| &lt; |x|$ 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:

$$|\sin(\sqrt{-1})| = 1.1752\ldots \not &lt; 1 = |\sqrt{-1}|$$

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).



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 $|\sin(x)| &lt; |x|$ (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)
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.