can we prove $\sin x<x$ using complex numbers

Question

In a book about complex numbers they have given the inequality $$\sin x\le x$$ for $x>0$

I know this to prove using differentiation and integration

my problem is can we prove this using theorems on complex numbers only

hints? thanks.

Answer 1

I don't know what kind of proof you are looking for, but I will try one proof which uses only the power series (and no differentiation, maximization etc). We have $\sin (x)= x -\{x^{2}/(2!)-x^{3}/3!\}-\{x^{4}/(4!)-x^{5}/5!\}-....$. So it suffices to check that $\{x^{2}/(2!)-x^{3}/3!\} \geq 0$, $\{x^{4}/(4!)-x^{5}/5!\} \geq 0$. All these are trivially true for $x<3$. So if you are willing to assume $\sin (x) \leq 1$ you get $\sin (x) \leq x$ for $x<3$ as well as $x \geq 3$.