18
$\begingroup$

I found a geometric method to show $$\text{when}\; x\to 0 \space , \space \cos x\sim 1-\frac{x^2}{2}$$ like below :

enter image description here

Suppose $R_{circle}=1 \to \overline{AB} =2$ in $\Delta AMB$ we have $$\overline{AM}^2=\overline{AB}\cdot\overline{AH} \tag{*}$$and $$\overline{AH}=\overline{OA}-\overline{OH}=1-\cos x$$ when $x$ is very small
w.r.t. * we can conclude $$x^2 \approx 2(1-\cos x) \\\to \cos x \sim 1- \frac{x^2}{2}$$ Now I have two question :
$\bf{1}:$ Is there other idea(s) to prove (except Taylor series) $x\to 0 , \space \cos x\sim 1-\frac{x^2}{2}\\$

$\bf{2}:$ How can show $\sin x \sim x- \frac{x^3}{6}$ with a geometric concept ?

Thanks in advance.

$\endgroup$
  • $\begingroup$ It's much more difficult, because $x$ is, in theory, a length of a curve, so $x-\sin x$ is comparing the arc MA to the length $MH$. $\endgroup$ – Thomas Andrews Aug 10 '17 at 20:07
  • 1
    $\begingroup$ $\cos x \sim 1-\frac{x^2}{2}$ is best interpreted as $$\lim_{x \to 0} \frac{1-\cos x}{\frac{x^2}{2}} = 1$$ In other words, it's stronger than just being "approximately" equal in a loose sense. I don't see how this geometric proof establishes the above limit. $\endgroup$ – MathematicsStudent1122 Aug 10 '17 at 20:24
  • $\begingroup$ Note that you obtained your result essentially by assuming that $\sin x = x + o(x^2)$ (by asserting that $\angle AOM \doteq HM$). Obviously, that won't help for demonstrating that $\sin x = x - \frac{x^3}{6} + o(x^3)$. $\endgroup$ – Brian Tung Aug 10 '17 at 20:24
  • $\begingroup$ This answer of mine shows a geometric interpretation of the power series of sine (and cosine, and secant and tangent) using iterated involutes. $\endgroup$ – Blue Aug 11 '17 at 2:32
  • $\begingroup$ Okay, I think I've got a geometric proof, using the areas rather than lengths. $\endgroup$ – Thomas Andrews Aug 11 '17 at 16:59
1
$\begingroup$

Here is a proof that uses the fact, once in $(4)$ and twice in $(5)$, that $$ \lim_{x\to0}\frac{\sin(x)}{x}=1\tag{1} $$ A geometric proof of $(1)$ can be found in this answer.


Answer Using the Duplication Formulas for $\boldsymbol{\sin(x)}$ and $\boldsymbol{\cos(x)}$

Using the duplication formulas for $\sin(x)$ and $\cos(x)$, we get $$ \begin{align} \frac{2\sin(x/2)-\sin(x)}{x^3} &=\frac{2\sin(x/2)-2\sin(x/2)\cos(x/2)}{x^3}\\ &=\frac{2\sin(x/2)}x\,\frac{1-\cos(x/2)}{x^2}\\ &=\frac{2\sin(x/2)}x\,\frac{2\sin^2(x/4)}{x^2}\tag{2} \end{align} $$ Substituting $x\mapsto\frac{x}{2^k}$ in $(2)$ gives $$ \frac{2^{k+1}\sin\left(\frac{x}{2^{k+1}}\right)-2^k\sin\left(\frac{x}{2^k}\right)}{x^3}=4^{-k}\frac{2\sin\left(\frac{x}{2^{k+1}}\right)}{\frac{x}{2^k}}\,\frac{2\sin^2\left(\frac{x}{2^{k+2}}\right)}{\frac{x^2}{4^k}}\tag{3} $$ Summing in $k$, the left side of $(3)$ telescopes: $$ \frac{x-\sin(x)}{x^3}=\sum_{k=0}^\infty4^{-k}\frac{2\sin\left(\frac{x}{2^{k+1}}\right)}{\frac{x}{2^k}}\,\frac{2\sin^2\left(\frac{x}{2^{k+2}}\right)}{\frac{x^2}{4^k}}\tag{4} $$ Since the terms of the right side of $(4)$ are dominated by $\frac18\cdot4^{-k}$, Dominated Convergence yields $$ \begin{align} \lim_{x\to0}\frac{x-\sin(x)}{x^3} &=\sum_{k=0}^\infty4^{-k}\lim_{x\to0}\frac{2\sin\left(\frac{x}{2^{k+1}}\right)}{\frac{x}{2^k}}\,\lim_{x\to0}\frac{2\sin^2\left(\frac{x}{2^{k+2}}\right)}{\frac{x^2}{4^k}}\\ &=\frac43\cdot1\cdot\frac18\\[6pt] &=\frac16\tag{5} \end{align} $$


Answer Inspired by Equation $\boldsymbol{(2)}$ from Ricardo Cruz's Answer

Recall the triplication formula for $\sin(x)$: $$ \sin(3x)=3\sin(x)-4\sin^3(x)\tag{6} $$ Substituting $x\mapsto\frac{x}{3^{k+1}}$ gives $$ \frac{3^{k+1}\sin\left(\frac{x}{3^{k+1}}\right)-3^k\sin\left(\frac{x}{3^k}\right)}{x^3}=4\cdot9^{-k}\frac{\sin^3\left(\frac{x}{3^{k+1}}\right)}{\frac{x^3}{27^k}}\tag{7} $$ Summing in $k$, the left side of $(7)$ telescopes: $$ \frac{x-\sin(x)}{x^3}=\sum_{k=0}^\infty4\cdot9^{-k}\frac{\sin^3\left(\frac{x}{3^{k+1}}\right)}{\frac{x^3}{27^k}}\tag{8} $$ Since the terms of the right side of $(8)$ are dominated by $\frac4{27}\cdot9^{-k}$, Dominated Convergence yields $$ \begin{align} \lim_{x\to0}\frac{x-\sin(x)}{x^3} &=\sum_{k=0}^\infty4\cdot9^{-k}\lim_{x\to0}\frac{\sin^3\left(\frac{x}{3^{k+1}}\right)}{\frac{x^3}{27^k}}\\ &=4\cdot\frac98\cdot\frac1{27}\\[6pt] &=\frac16\tag{9} \end{align} $$

$\endgroup$
5
$\begingroup$

Let assume that for small $x$, $$\sin x = x + a x^2 + bx^3 + O(x^4)\tag{1}$$

enter image description here

In this trigonometric circle we can compute $BC$ in two ways:

First, considering the $ABC$ triangle:

$$BC= \sqrt{AB^2+AC^2}=\sqrt{\sin^2 x+\left(1-\sqrt{1-\sin^2 x}\right)^2}=\\= x+ax^2+\left(b+\frac18\right)x^3+O(x^4) \tag{2}$$ [*]

Second, bisecting the angle $x$:

$$BC = 2 \sin \frac{x}{2}=x+\frac{a}{2} x^2 + \frac{b}{4}x^3 + O(x^4)\tag{3}$$

Equating (2) and (3) we get $a=0$ and $b=-\frac{1}{6}$, that is: $\sin x = x - \frac16 x^3 + O(x^4)$

I'm not sure if this a geometric proof, though...


[*] To prove (2) we only need to know the expansion, for $t\to 0$: $$\sqrt{1+t}=1+\frac{t}{2}+\cdots \tag{4}$$ (here and in the following expansions the dots means a remainder $O(t^{k+1})$ where $k$ is the highest exponent in the preceeding terms).

Then $$\sqrt{1-t^2}=1-\frac{t^2}{2}+\cdots \\ (1-\sqrt{1-t^2})^2=\frac{t^4}{4}+\cdots\\ t^2+(1-\sqrt{1-t^2})^2=t^2(1+\frac{t^2}{4}+\dots)\\ \sqrt{t^2+(1-\sqrt{1-t^2})^2}=t (1+\frac{t^2}{8}+\cdots) =t +\frac{t^3}{8}+\cdots $$ Letting $t=x+ax^2+bx^3+O(x^4)$ that gives

$$ x+ax^2+bx^3+\frac{x^3}{8}+O(x^4)$$

With some more work (I've verified it with Maxima), doing higher order expansions we can get more terms ($\sin x = x -\frac16 x^3 + \frac{1}{120}x^5 +\cdots$).

$\endgroup$
  • $\begingroup$ can you elaborate a little more about your methods ? $\endgroup$ – Khosrotash Aug 11 '17 at 3:58
  • $\begingroup$ I don't understand right hand side of part (2) ...can you elaborate more ? $\endgroup$ – Khosrotash Aug 11 '17 at 11:46
  • $\begingroup$ It's just a Taylor expansion. $\endgroup$ – leonbloy Aug 11 '17 at 13:46
  • $\begingroup$ Since we know $\sin(-x)=-\sin(x)$, you know $a=0$. $\endgroup$ – Thomas Andrews Aug 11 '17 at 15:33
  • $\begingroup$ The problem is that all you can say is that $\sin^2(x)=x^2+O(x^6)$. You need to use that $\sin(-x)=-\sin(x)$ to assume that $\sin(x)=x+bx^3+O(x^5)$. $\endgroup$ – Thomas Andrews Aug 11 '17 at 15:35
1
$\begingroup$

Use the formula: $$\sin(a+b)=\sin(a) \cos (b)+\sin(b) \cos (a) \quad (1)$$ Which has a geometrical derivation, to find out $\sin(x)$ in function of $\sin(\frac{x}{3})$. That produces the triple angle formula: $$\sin(x)=3\sin(\frac{x}{3})-4(\sin(\frac{x}{3}))^3 \quad(2)$$ Using the approximation: $$\sin(\frac{x}{9})=\frac{x}{9} \quad (3)$$ And (2) twice, first to get $\sin(\frac{x}{3})$ and after that to get $\sin(x)$, you will get a polynomial: $$\sin(x)=x-\frac{40}{243}x^3 + \text {ninth degree polynomial} \quad (4)$$ The polynomial can be approximated to: $$\sin(x)=x-\frac{1}{6}x^3 $$.

$\endgroup$
  • $\begingroup$ can you explain more ? specially (3) $\endgroup$ – Khosrotash Aug 16 '17 at 13:08
  • $\begingroup$ @Khosrotash I assume that $\sin(\frac{x}{9}) \sim \frac{x}{9}$ is a good approximation. Then I use $\sin(\frac{x}{3})=3\sin(\frac{x}{9})-4(\sin(\frac{x}{9}))^3$ to get: $\sin(\frac{x}{3}) \sim 3(\frac{x}{9})-4(\frac{x^3}{729})$. After that I substitute the last expression into (2), obtaining (4). $\endgroup$ – RicardoCruz Aug 16 '17 at 13:24
  • $\begingroup$ Why $\frac x9$ ? $\endgroup$ – Khosrotash Aug 16 '17 at 13:28
  • $\begingroup$ @Khosrotash Because dividing an arch in thirds I can use the sine recurrently, without the necessity of cosine. $\endgroup$ – RicardoCruz Aug 16 '17 at 13:33
  • $\begingroup$ @Khosrotash I posted another answer with a different approach. This new approach uses the area of circular segment. $\endgroup$ – RicardoCruz Aug 16 '17 at 14:01
1
$\begingroup$

This answer uses another approach. We can approximate the area of a circular segment to an area of a parable. See the picture below.

ParableCircleApp

The area of a circular segment can be calculated by: $$A=\frac{R^2}{2}(\theta-\sin(\theta)) \tag{1}$$

The area of a parable between its zeroes $x_1$ and $x_2$ can be calculated by: $$A=\frac{2}{3}(x_2-x_1)y_{max} \tag{2}$$ where $y_{max}$ is the maximum ordinate of the parable.

If we use: $$x_2 =\sin(\frac{\theta}{2}) \tag{3}$$ $$x_1 =-\sin(\frac{\theta}{2}) \tag{4}$$ $$y_{max}=1-\cos(\frac{\theta}{2})= 2 (\sin(\frac{\theta}{4}))^2 \tag{5}$$ $$\sin(\frac{\theta}{2}) \sim \frac{\theta}{2} \tag{6}$$ $$\sin(\frac{\theta}{4}) \sim \frac{\theta}{4} \tag{7}$$ and $$R=1 \tag{8}$$ in equations (1) and (2) and after that we make the approximation that (1) and (2) are the same, we get: $$\sin(\theta) \sim \theta - \frac{\theta^3}{6}.$$

$\endgroup$
  • 1
    $\begingroup$ That's slick, but doesn't show that the difference between the two areas is $o(\theta^3)$, which is necessary for $\sim $. $\endgroup$ – Thomas Andrews Aug 16 '17 at 15:15
  • $\begingroup$ Can we find the area under a parabola without using calculus? $\endgroup$ – robjohn Aug 19 '17 at 13:27
  • $\begingroup$ @robjohn Archimedes calculated the area using the method of exhaustion in his work The quadrature of the parabola, which is similar to modern calculus. $\endgroup$ – RicardoCruz Aug 19 '17 at 16:42
1
$\begingroup$

Now a full answer (assuming we have a result slightly stronger than yours, that $\cos x = 1-\frac{x^2}{2}+o(x^2)$, which is the true meaning of $\cos x\sim 1-\frac{x^2}{2}$, but which you haven't proven - you haven't come up with a bound for your error.)

Let $D$ be the midpoint of the arc in $MA$. The area of the quadrilateral $OMDA$ is $\frac{\sin(x)}{2}$.

The area of circular wedge between $M$ and $A$ is $\frac{x}{2}$.

Taking the tangents to the circle at $M$ and $A$ and finding their intersection, $C$, you get that the are of $OMCA$ is $2\tan(x/2)$. So you have:

$$\sin\frac{x}{2}<\frac{x}{2}<\tan\frac{x}{2}$$

Setting $u=\frac{x}{2}$ you get:

$$\sin u< u < \frac{\sin u}{\cos u}$$

Or $\cos u<\frac{\sin u}{u}<1$

This is enough to see that $\sin u = u+O(u^3)$ (using that $\cos(u)=1+O(u^2)$,) but not enough to show what you want.

The goal is to find and approximation of the difference $u-\sin u$, twice the area of the region between the line segment $AD$ and the arc $AD$.


Letting $$A_n=\frac{1}2\sin \frac{x}{2^{n}},$$ is the area of $2^{n}$ copies of the triangle $MOA$ with angle $\frac{x}{2^n}$. From this, we see that $2^{n}A_n<\frac{x}{2}$, since the $2^n$ triangles fit inside the wedge.

Now, $A_{n+1}$ is gotten by taking half of the triangle for $A_{n+1}$ and adding a triangle with base $\sin(x/2^{n+1})$ and height $1-\cos(x/2^{n+1})$.

So: $$A_{n+1}=\frac{A_n}{2}+\frac{1}{2}\sin(x/2^{n+1})(1-\cos(x/2^{n+1})).$$

So if $B_{n}=2^{n}A_n$, then $B_0=\frac{\sin x}{2}$ and $$B_{n+1}=B_n+2^{n}\sin(x/2^{n+1})(1-\cos(x/2^{n+1})).$$

Writing $f(x)=x-\sin x$ and $g(x)=\cos x-1+\frac{x^2}{2}$ as our error functions. We get:

$$\begin{align}B_{n+1}-B_{n}&=2^{n}\sin(x/2^{n+1})(1-\cos(x/2^{n+1}))\\&=2^{n}\left(\frac{x}{2^{n+1}}+f\left(\frac{x}{2^{n+1}}\right)\right) \left(\frac{x^2}{2^{2n+3}}+g\left(\frac{x}{2^{n+1}}\right)\right)\\ &=\frac{x^3}{2^{2n+4}}+\frac{x}{2}g\left(\frac{x}{2^{n+1}}\right)+\frac{x^2}{2^{n+3}}f\left(\frac{x}{2^{n+1}}\right)+f\left(\frac{x}{2^{n+1}}\right)g\left(\frac{x}{2^{n+1}}\right) \end{align}$$

Now, we know that $f(x)=O(x^3)$ and $g(x)=o(x^2)$, so we get:

$$\begin{align}\lim_{n\to\infty} B_n &= B_0+\sum_{n=0}^{\infty}\left(B_{n+1}-B_n\right)\\ &=\frac{\sin x}{2}+\left(x^3\sum_{n=0}^{\infty}\frac{1}{2^{2n+4}}\right)+o(x^3)+O(x^5)+o(x^5)\\ &=\frac{\sin x}{2}+\frac{x^3}{12}+o(x^3) \end{align}$$

But $\lim_{n\to\infty} B_n =\frac{x}{2}$, because this is the area of the circular wedge.

So we have: $$x-\sin x = \frac{x^3}{6} + o(x^3)$$


You can actually prove that $B_n\to \frac{x}{2}$ by noting that $\frac{B_{n}}{\cos(x/2^n)}$ is the area of $2^n$ triangles that cover the circular wedge, and thus you have:

$$B_n<\frac{x}{2}<\frac{B_n}{\cos(x/2^n)}$$

Simce $B_n$ is increasing, and $\cos(x/2^n)\to 1$, we see that both values converge to $\frac{x}{2}$.


The full proof first prove that $\sin x = x+o(x)$.

From this, though, we can show that:

$$\cos(x)=1-2\sin^2 \frac{x}{2} = 1-\frac{x^2}{2}+o(x^2).$$

Then the rest of the argument follows, since $\cos(x)=1+O(u^2)$.

$\endgroup$
  • $\begingroup$ Let $S$ be the point on line $\overline{MH}$ such that the perpendicular line through $S$ hits $D$. Then $\overline{MH}=\overline{MS}+\overline{SH}$. I'm lazy to work it out, but wouldn't you get a better approximation of $\overline{MH}$ this way by expressing the two segments in terms of $x/2$ (using pythagorean theorem)? $\endgroup$ – Alex R. Aug 10 '17 at 20:32
  • $\begingroup$ The proof can likely use an outer limit area, rather than "knowing" that $\lim_{n\to \infty} B_n=\frac{x}{2}$, you can use that $B_n <\frac{x}{2}<C_n$ where $C_n=2^{n}\tan(x/2^{n+1})$. Then if you show that $\lim C_n =\frac{ \sin x}{2}+ \frac{x^3}{12}+o(x^3)$ you have that $x=\sin x + \frac{x^3}{6}+o(x^3)$. $\endgroup$ – Thomas Andrews Aug 11 '17 at 17:03
0
$\begingroup$

Not a complete answer, I am afraid. Approximating the arc $AB$ as a line segment leads to the correct approximation of $\cos(x) \sim 1 - \frac{x^2}{2}$ but leads to the inaccurate result for the corresponding sine as $ \sin(x) \sim x - \frac{x^3}{8}$. Nevertheless, I post my approach, since it has not been covered in the preceding answers.

enter image description here

In the above diagram, we have $OA=OB=1$ and $\angle BOA = x$

Let the co-ordinates of point $B$ be $(h,k)$

Since $B$ lies on the unit circle, $$h^2 + k^2 =1\tag{1}$$ The equation of the line $AB$ is given as $y^*=m(x^*-1)$ and since B lies on the line $AB$, we have $$ k=m(h-1\tag{2})$$ where $m$ is the slope of $AB$.

From $(1)$ and $(2)$, we have $$\Rightarrow h^2 + m^2(h-1)^2=1 $$ $$\Rightarrow h^2(1+m^2) - h(2m^2) + (m^2 -1)=0 $$ which gives two values of $h$, of which one value $h=1$ is for the point $A$ $$\Rightarrow h=\frac{m^2 -1}{m^2 +1}, \; k =\frac{-2m}{m^2 +1} \tag{3}$$

The length of the line segment $AB$ is assumed to be $x$, hence $$\Rightarrow (h-1)^2 + k^2 =x^2 \tag{4}$$

Using $(2)$ and $(4)$, we get $$ \Rightarrow (1+m^2)(h-1)^2 =x^2$$ $$\Rightarrow x=|h-1|\sqrt{1+m^2}=(1-h)\sqrt{1+m^2}$$ From $(3)$, we have $$ \Rightarrow x=\frac{2}{\sqrt{1+m^2}}$$ $$ \Rightarrow 1+m^2 = \frac{4}{x^2}$$ $$ \Rightarrow m = \sqrt{\frac{4-x^2}{x^2}} \tag{5}$$

Now, using $(5)$ and $(3)$, $$ h=OD=\frac{4-2x^2}{4}, \; k=DB= \frac{x\sqrt{4-x^2}}{2}$$

Thus $OD = \cos(x) \sim 1-\frac{x^2}{2}$ and $DB= \sin(x) \sim x(1-\frac{x^2}{4})^{1/2}$

This gives $\sin(x) \sim x - \frac{x^3}{8}$, which isn't quite correct.

$\endgroup$
  • $\begingroup$ "The length of the line segment AB is assumed to be x" That is wrong. $\endgroup$ – leonbloy Aug 15 '17 at 22:38

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.