I'm stuck with this proof. Please help.
Note: I can't use integration, only derivatives, taylor, etc.
Prove $\sin(x)<x$, $\forall x>0$
I try this:
$\sin(x)=\sum_{n=0}^{\infty}\;(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$ Then we need prove this $\sum_{n=0}^{\infty}\;(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}<x$... Well, I try to make this by induction but I'm stuck when I go to prove for $k+1$, Can someone help me?
Define for $\;x\ge 0\;$:
$$f(x)=\sin x-x\implies f'(x)=\cos x-1\le0\implies f\;\text{ is monotone descending}\implies$$
$$\forall\,x>0\;,\;\;f(x)\le f(0)=0\implies\;\text{we're done.}\;\;\;\;\square$$
If you need strict inequality just show, or even just mention (depending on what you know) that the function doesn't vanish in any non-trivial interval.
We know that $\sin x\le 1$, which answers the question for $x>1$.
For $0<x\le 1$ use the theorem about alternating series and the first three terms of the sinus series to get $$ \sin x\le x-\frac16x^3+\frac1{120}x^5=x-\frac1{120}x^3(20-x^2)<x $$ which is valid for $|x|<\sqrt{20}$.
If $f(x)=\sin x-x$, then $f'(x)=\cos x-1\le0$, so the function is decreasing. However, in a punctured neighborhood of $0$, we have $\cos x<1$, by using the fact that the series for the cosine is alternating.
If you don't know that $\cos x\le1$, then prove $\sin^2x+\cos^2x=1$, by differentiating the left-hand side.
Let $x>0$.
if $x>1$ then $\sin(x)\leq1<x$.
assume now that $0<x\leq1$
the sinus function is continuous at
$[0,x]$ and has a derivative at $(0,x)$
thus
$\exists c \in (0,x) : \sin(x)=x\cos(c)$.
$0<c<x\leq1 \implies$
$ 0<\cos(c)<1 \implies$
$\sin(x)<x$.
A proof with geometric color:
Let $x > 0$. Note that $x > 1$ implies $\sin x < x$ by the theorem that $\sin x \leq 1$ for all $x \in \mathbb{R}$. Suppose $0 < x \leq 1$. Note that $0 < x < \pi/2$ implies $\sin x < x$ by interpreting $x$ as the arc length of the unit circle under consideration. So $0 < x \leq 1$ implies $\sin x < x$.
