Ytukyg

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

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 $sin x$?

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 $x>0$.

Define the function $f(x)$ as

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

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 $xge 0$ with $f(x)>x$ for $0<xle 1$.

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

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

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

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

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

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.

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.

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

Theorem: Assuming the above statement, we have $sin(aepsilon)le aepsilon$ for all $age1$.

We only need to prove it for $aepsilonle1$, because we know that $sin 1le1$ (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 $0<aepsilon<1$, we have:
$$0<cos(aepsilon)le1, 0<sin(epsilon)leepsilon$$
and$$0<sin(aepsilon)le aepsilon, 0<cos(epsilon)le1$$
(The inequalities with sine follow from the hypothesis and the induction hypothesis.)

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

Adding them together:
$$cos(aepsilon)sin(epsilon)+sin(aepsilon)cos(epsilon)le(a+1)epsilon$$
$$sin((a+1)epsilon)le(a+1)epsilon$$
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, $sin epsilonapproxepsilon$ 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 $sinepsilon=epsilon$. Now, because $epsilon$ is infinitesimal, every real number $x$ is a multiple of it. So, using the above theorem, we now have $sin xle x$ for all positive $x$. (A sketchy) QED.

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

"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 $$x^{2} + y^{2} = 1tag{1}$$ 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 $$sin L < Ltag{2}$$ 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. $$pi = 2int_{0}^{1}frac{dx}{sqrt{1 - x^{2}}}$$ 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.