# how to strictly prove $\sin x<x$ for $0<x<\frac{\pi}{2}$

$$\sin x<x\,(0<x<\frac{\pi}{2})$$ In most textbook,to prove this inequality is based on geometry illustration(draw a circle, compare arc length and chord ),but I think that strict prove should based on analysis reasoning without geometry illustration.Who can prove it?Thank you very much.

ps:

1.By differentiation ,monotonicity and Taylor formula ,all are wrong,because $(\sin x)'=\cos x$ must use $\lim_{x \to 0}\frac{\sin x}{x}=1$,and this formula must use $\sin x< x$.This is vicious circle.

2.If we use Taylor series of $\sin x$ to define $\sin x$,strictly prove $\sin x<x$ is very easy,but how can we obtain geometry meaning of $\sin x$?

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You don't $have$ to define $\sin{x}$ in such a way to make that definition circular... for example, you could define it as a Taylor Series. –  Tyler Mar 28 '12 at 2:08
Please don't use displayed math in titles. –  Arturo Magidin Mar 28 '12 at 2:11
@TylerBailey I can define sin as a Taylor Series,but how to reason geometry interpretation of $\sin x$. –  noname1014 Mar 28 '12 at 2:13
Rudin's Principles of Mathematical Analysis (PMA) will be a good reference to the approach you're searching for. It begins with Taylor series to define sine and cosine, and deduce its properties purely out of it. For example differentiating the expression $$\left[\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{2n}\right]^2 + \left[\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}x^{2n+1}\right]^2$$ yields 0 identically, so we can deduce that it is identically 1. –  sos440 Mar 28 '12 at 2:16
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Define the function $f(x)=x-\sin x.$ Observe that $f(0)=0$ and $f'(x)=1-\cos x \geq 0$. The derivative is equal to $0$ only at isolated points, so the function increases in the interval $[0, \infty)$. That is, for all $x>0$ we have $f(x)>f(0)=0$. Thus $x>\sin x$ for all $x>0$.

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please read ps. –  noname1014 Mar 28 '12 at 8:06

We can define $\sin x$ as power series .Applying the knowledge of power series,obtain the derivative of $\sin x$,and then we will easy prove the inequality.Concluding geometry of $\sin x$ , please refer to this.

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Here is what a while ago I wrote based on a caculus challenge problem. It can be seen here.

Here is another one.

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please read ps. –  noname1014 Mar 28 '12 at 8:05
I updated my answer. –  Vafa Khalighi Mar 29 '12 at 0:18
thnks，I think define sinx by Taylor series ,it will be easy to prove sinx < x,the rest problem is how to find geometry meaning of sinx.math.stackexchange.com/questions/125511/… –  noname1014 Mar 29 '12 at 0:29
Assuming $\epsilon$ to be a very small and nearly zero in value, the area of $sin(x) in the desired interval is approximately is$A1=\int_{0+\epsilon }^{\pi/2 - \epsilon }sin(x)dx = cos(0+\epsilon )-cos(\pi/2 - \epsilon ) \approx cos(0)-sin(\epsilon )\approx 1$The area under the line$y=x$for the same interval is:$A2=\int_{0+\epsilon }^{\pi/2 - \epsilon} x dx = \frac{1}{2}(\pi/2-\epsilon )^2 - \epsilon ^2) \approx1.23$Since A1 < A2, we can say that:$\sin x<x\,(0<x<\frac{\pi}{2})\$