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


  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. – Sangchul Lee Mar 28 '12 at 2:16

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
up vote 3 down vote accepted

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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Define the function $f$ as $$f(x) = \int_0^x\frac{1}{\sqrt{1 - t^2}} \,dt$$

as the inverse function for the sine function. Using standard analysis (e.g. L'Hospital's rule), and without appeal to any a priori knowledge of properties of $\sin (x)$ and $\cos (x)$, one can derive easily the limit sought.

(1) $f(0) = 0$ implies $\sin (0) = 0$ since $f(x)$ is the inverse function of $\sin (x)$;

(2) $\frac{df(x)}{dx} = \frac{1}{\sqrt{1 - x^2}}$ follows from the fundamental theorem of calculus;

(3) $\frac{d\sin (x)}{dx} = \sqrt{1 - \sin^2 (x)}$ using (2) along with the relationship between derivatives of inverse functions;

(4) $\lim_{x\to 0} \frac{\sin (x)}{x} = \lim_{x\to 0} \frac{d \sin (x)}{dx}$ follows from L'Hospital's Rule;

(5) Using (1)-(4), $\lim_{x\to 0} \frac{\sin (x)}{x} =\lim_{x\to 0} \frac{d \sin (x)}{dx} = \lim_{x\to 0} \sqrt {1 - \sin^2 x} = 1$

which completes the proof.

And this proof did not appeal to any prior knowledge of the sine function and did not even mention the cosine function!

This is probably the simplest, rigorous proof available.

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May be you can prove the fact by finding the area under the curve of each function.

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:

$\sin x<x\,(0<x<\frac{\pi}{2})$

enter image description here

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Assume, for some $\epsilon$ small enough, we have $\sin(\epsilon)\le\epsilon$. (I'll deal with this statement later.)

Theorem: Assuming the above statement, we have $\sin(a\epsilon)\le a\epsilon$ for all $a\ge1$.

We only need to prove it for $a\epsilon\le1$, because we know that $\sin 1\le1$ (because $\sin$ is always less then 1). We proceed by induction. We know (assuming the above statement) that it's true for $a=1$. Now, assume it's true for $a$; we need to prove it for $a+1$.

Note that, for $0<a\epsilon<1$, we have: $$0<\cos(a\epsilon)\le1,\ 0<\sin(\epsilon)\le\epsilon$$ and$$0<\sin(a\epsilon)\le a\epsilon,\ 0<\cos(\epsilon)\le1$$ (The inequalities with sine follow from the hypothesis and the induction hypothesis.)

Multiplying them together, we have: $$\cos(a\epsilon)\sin(\epsilon)\le\epsilon$$ $$\sin(a\epsilon)\cos(\epsilon)\le a\epsilon$$ (We needed to know that they were positive, because then we know we don't have to switch around the inequality.)

Adding them together: $$\cos(a\epsilon)\sin(\epsilon)+\sin(a\epsilon)\cos(\epsilon)\le(a+1)\epsilon$$ $$\sin((a+1)\epsilon)\le(a+1)\epsilon$$ where I used the sum formula for sine in the last line. QED.

Now, here I'm going to have to use some sketchiness. Recall how, with radians, $\sin \epsilon\approx\epsilon$ when $\epsilon$ is small. Thus, if we let $\epsilon$ be an infinitesimal number (I told you I'm going to have to use some sketchiness), we basically have $\sin\epsilon=\epsilon$. Now, because $\epsilon$ is infinitesimal, every real number $x$ is a multiple of it. So, using the above theorem, we now have $\sin x\le x$ for all positive $x$. (A sketchy) QED.

If anything in this comment is incoherent, I apologize—I am currently extremely tired.

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