How to prove the Jordan's Inequality?

Question

Can anyone tell me how to prove it by using the concepts related to limit?

Answer 1

Notice that |sin(x)|=sin(|x|) and without loss of generality assume that $0\leq x\leq \frac{\pi}2$. Let $x_0=\arccos(\frac{2}\pi)$ so that for $x\leq x_0$ we have $\cos(x)\geq \frac{2}\pi$ and thus, $$ \sin(x)=\displaystyle\int_{0}^{x} \cos(y)dy \geq \displaystyle\int_{0}^{x}\frac{2}\pi dy=\frac{2}\pi x $$ while for $x\geq x_0$ we have $\cos(x)\leq \frac{2}\pi$ and thus, $$ \sin(x)=1-\displaystyle\int_{x}^{\frac{\pi}2 } \cos(y)dy \geq 1-\displaystyle\int_{x}^{\frac{\pi}2 }\frac{2}\pi dy=1-\frac{2}\pi(\frac{\pi}2-x)=\frac{2}\pi x $$

NB: the nicest way to do it is as @Etienne said is the concavity of sin: The sine function is concave on $[0,\frac{\pi}2]$ because $f '(x) = \cos (x)$ is decreasing on $[0,\frac{\pi}2]$.

Thus, $$\underbrace{\sin( x)}_{\sin \text{curve}} \geq \overbrace{\dfrac{2}{\pi} x}^{\text{line joining 0 and $\pi$/2}}$$