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

  • 3
    $\begingroup$ 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. $\endgroup$
    – Tyler
    Mar 28, 2012 at 2:08
  • 2
    $\begingroup$ mathrefresher.blogspot.com/2006/08/… $\endgroup$ Mar 28, 2012 at 2:10
  • $\begingroup$ Please don't use displayed math in titles. $\endgroup$ Mar 28, 2012 at 2:11
  • $\begingroup$ @TylerBailey I can define sin as a Taylor Series,but how to reason geometry interpretation of $\sin x$. $\endgroup$
    – noname1014
    Mar 28, 2012 at 2:13
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 28, 2012 at 2:16

6 Answers 6


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 $x>0$.

  • 2
    $\begingroup$ please read ps. $\endgroup$
    – noname1014
    Mar 28, 2012 at 8:06

Define the function $f(x)$ as

$$f(x) = \int_0^x\frac{1}{\sqrt{1 - t^2}} \,dt \tag 1$$

for $|x|\le 1$, as the inverse function for the sine function. That is to say that $\sin(x)=f^{(-1)}(x)$.

Clearly, $f(x)$ as given in $(1)$ is monotonically increasing with $0=f(0)\le f(x)\le f(1)=\pi/2$. Furthermore, it is easy to see from $(1)$ that $f(x)\ge x$ for $x\ge 0$ with $f(x)>x$ for $0<x\le 1$.

Inasmuch as $f$ is monotonically increasing on $(0,1)$, its inverse, $\sin(x)$, is monotonically increasing with $0\le \sin(x)\le 1$ for $0\le x\le \pi/2$. Moreover, we see that since $f(x)>x$ for $0<x\le 1$, then $0< \sin(x) <x$ for $0 <x\le \pi/2$.

EVALUTING THE LIMIT $\displaystyle \lim_{x\to 0}\frac{\sin(x)}{x}$

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.

  • 2
    $\begingroup$ Shame I've never seen this proof before! $\endgroup$ May 31, 2016 at 6:32

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.


"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."

The geometrical figure accompanying such a proof is an illustration to help you understand the argument. The proof is not dependent on the geometrical illustration but is based on certain basic facts about arc-length.

The concept of arc length itself is based on analytic notions of supremum and the fact that an arc of a circle has a length is dependent on the fact that the function $f(x) = \sqrt{1 - x^{2}}$ is monotone on $[0, 1]$. Thus I remove the geometrical illustration here (i.e. I remove the drawing and you will have to spend extra effort to visualize / understand what follows) and present the classical proof.

Let $\mathcal{C}$ be the set of points $(x, y)$ in plane which satisfy $$x^{2} + y^{2} = 1\tag{1}$$ and consider a part of it namely the smaller part from point $A = (1, 0)$ to $B = (0, 1)$ and let $P = (p, q)$ be an intermediate point on this arc $AB$. Let $C = (p, 0)$ and clearly we can see that the length of line segment $PC$ i.e. $q$ is less than the length of line segment $AP$ i.e. $\sqrt{(p - 1)^{2} + q^{2}}$.

Now the calculation of arc length $L$ of arc $AP$ is somewhat tedious. If is defined as the supremum of lengths of a polygonal arcs with points lying on arc $AP$ such that $A$ is the starting point of the polygonal arcs and the final point is $P$. Clearly line segment $AP$ of length $\sqrt{(p - 1)^{2} + q^{2}}$ is also one such polygonal arc and hence by definition of supremum we have $$L \geq \sqrt{(p - 1)^{2} + q^{2}} > q$$ By the definition of circular functions the coordinates of point $P$ are $(\cos L, \sin L)$ and hence $p = \cos L, q = \sin L$. We have thus proved that $$\sin L < L\tag{2}$$ for values of $L$ such that the the point $P$ lies in the smaller part of the curve $\mathcal{C}$ between $A$ and $B$. This means that $L > 0$ and $$L < \int_{0}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$$ where the integral on right is the length of the whole arc $AB$ based on formula of arc-length in terms of integral. If we define circular functions on the basis of arc-length (as done above) then the constant $\pi$ is defined to be twice the above integral i.e. $$\pi = 2\int_{0}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$$ Thus we have finally proved that $\sin L < L$ for $0 < L < \pi/2$.

Contrary to what many believe the definition of circular functions via the use of circle $\mathcal{C}$ defined by $x^{2} + y^{2} = 1$ is fully rigorous. This is also the traditional approach used in textbooks of trigonometry. However the fundamental point in this approach is to justify one of the following two things and define the circular functions accordingly:

  1. An arc of a circle has a well defined length. If $A = (1, 0)$ is a fixed point on circle $\mathcal{C}$ and $P$ is a point on same curve $\mathcal{C}$ with length of arc $AP$ as $x$ then the coordinates of the point $P$ are defined to be $(\cos x, \sin x)$.
  2. A sector of a circle has a well defined area. If $A = (1, 0)$ is a fixed point on circle $\mathcal{C}$ and $P$ is a point on same curve $\mathcal{C}$ with area of sector corresponding to arc $AP$ as $x/2$ then the coordinates of the point $P$ are defined to be $(\cos x, \sin x)$.

The traditional presentation of circular functions uses one of the two above definitions (the one on area is simpler to execute) but typically the justifications of length and area are ommitted because they require the notion of integral. Moreover it can be easily established via integral calculus that both the above definitions are equivalent. For more details see my blog post.

  • $\begingroup$ We can use this instead of dyadic rationals to get the $sin(x) / x$ fix. $\endgroup$ Jun 21, 2017 at 12:03
  • $\begingroup$ @MikeMathMan: quite true! I prefer this approach because many students are already used to it and the only thing needed to make it rigorous is the justification of area and length. But your approach via dyadic rationals is also nice. The argument to move from dyadic rationals to reals is well known and should not be difficult to grasp for many. $\endgroup$
    – Paramanand Singh
    Jun 21, 2017 at 13:58
  • $\begingroup$ It wasn't me who put in the 'proposed fix' - it was @jack-daurizio who argued that I was hand waving (and he was correct). $\endgroup$ Jun 21, 2017 at 14:03
  • $\begingroup$ @MikeMathMan: oh, perhaps you should indicate in your answer something like "proposed fix by Jack d'Aurizio". $\endgroup$
    – Paramanand Singh
    Jun 21, 2017 at 15:00

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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.