Elementary proof (involving no series expansion) that $\tan x \approx x+\frac{x^3}{3}$ for small $x$

Question

For small values of $x$, the following widely used approximations follow immediately by Taylor expansion:

1. $\sin x \approx x$
2. $\cos x \approx 1-\frac{x^2}{2}$
3. $\tan x \approx x +\frac{x^3}{3}$

I am looking for a justification of these approximations without the use of series expansions.

By purely geometric considerations, it is easy to see that for small values of $x$, we have

$$ \sin x < x < \tan x.$$

Division by $\cos x$ and an application of the squeeze lemma yield

$$\frac{\sin x}{x}\xrightarrow{x\to0}1$$

and hence the approximation (1.). Using the identity $\cos x = 1-2 \sin^2 \frac{x}{2}$, the approximation (2.) also follows.

Can one justify the approximation (3.) by a similarly elementary argument without using Taylor expansion?

I tried around using the angle addition theorems, but I did not really get anywhere, mainly because I could not make the factor $\frac13$ appear anywhere.

Answer 1

Famously, a sector with a small angle proves $\lim_{x\to0}\tfrac{\sin x}{x}=1$ and hence$$\lim_{x\to0}\tfrac{1-\cos x}{x^2}=\lim_{x\to0}\tfrac{\sin^2\tfrac{x}{2}}{2(x/2)^2}=\tfrac12.$$We can improve the former result: if you approximate a sector's arc as a parabola, you can prove$$\lim_{x\to0}\frac{1-\tfrac{\sin x}{x}}{x^2}=\tfrac16.$$Finally,$$\lim_{x\to0}\frac{\tfrac{\tan x}{x}-1}{x^2}=\lim_{x\to0}\frac{\tfrac{\sin x}{x}-\cos x}{x^2}=\lim_{x\to0}\frac{1-\cos x}{x^2}-\lim_{x\to0}\frac{1-\tfrac{\sin x}{x}}{x^2}=\tfrac12-\tfrac16=\tfrac13.$$

Answer 2

Look at the trigonometric circle and approximate the chord $s$ with the angle $\theta$.

Here $x=\cos \theta$ and $y=\sin \theta$.

From the big triangle, you have $x^2+y^2=1$ which is obvious, and from the small circle, you have $s^2 = y^2 + (1-x)^2$. But with approximation $s \approx \theta$ the above is

$$\theta^2 \approx y^2 + (1-x)^2 \tag{1}$$

Now sub $y=\sqrt{1-x^2}$ to get the well-known approximation

$$ \theta^2 \approx 1-x^2 + (1-x)^2 = 2 (1-x) $$

or

$$ x = \cos \theta \approx 1 - \tfrac{1}{2} \theta^2 \tag{2}$$

Now use (2) in (1) to get $\theta^2 \approx y^2 + \tfrac{1}{4} \theta^4$, but since we are keeping this a second order approximation $\theta^4 \approx 0$ and

$$ y = \sin \theta \approx \theta \tag{3} $$

Finally, $\tan \theta = \tfrac{y}{x} \approx \tfrac{\theta}{1-\tfrac{1}{2} \theta^2}$. But use the property $\tfrac{1}{1-z} = 1 + \frac{1}{1-z} z \approx 1 + z$ with $z=\tfrac{1}{2} \theta^2$ to get the final approximation.

$$ \tfrac{y}{x} = \tan \theta \approx \theta \left( 1 + \tfrac{1}{2} \theta^2 \right) \tag{4a} $$

But note that the last one is based on two approximations.

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Another approach with $\tan$ would be to approximate once $\tan \theta = \tfrac{y}{x} = \theta \frac{\sqrt{1-\tfrac{1}{2}z}}{1-z}$ with $z=\tfrac{1}{2}\theta^2$ and the approximation $ \tan \theta \approx \theta \left( 1 + \tfrac{3}{4} z \right)$

$$ \tfrac{y}{x} = \tan \theta \approx \theta \left( 1 + \tfrac{3}{8} \theta^2 \right) \tag{4b} $$

For example with $\theta=0.6$ you have $\tan(0.6) = 0.68413681$, $0.6 \left( 1 + \tfrac{1}{2} 0.6^2 \right)=0.708$ an 3.5% error, and $0.6 \left( 1 + \tfrac{3}{8} 0.6^2 \right)=0.681$ an 0.45% error.

So (4b) is at least an order of magnitude better at approximating $\tan \theta$ than (4a). A graphical comparison of the error $f(x)/\tan x-1$ is shown below:

Answer 3

Proof that $\boldsymbol{\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}}=\frac16$

As shown in $(2)$ from this answer, since $\cos(x)$ is continuous, given any $\epsilon\gt0$, we can find a $\delta\gt0$ so that if $0\lt|x|\le\delta$, we have $$ 1-\epsilon\le\cos(x)\le\frac{\sin(x)}x\le1\tag1 $$ Thus, assuming $0\lt|x|\le\delta$, $$ \begin{align} \frac{x-\sin(x)}{x^3} &=\frac1{x^3}\sum_{n=0}^\infty\left(2^{n+1}\sin\left(x/2^{n+1}\right)-2^n\sin\left(x/2^n\right)\right)\tag{2a}\\ &=\frac1{x^3}\sum_{n=0}^\infty\left(2^{n+1}\sin\left(x/2^{n+1}\right)-2^{n+1}\sin\left(x/2^{n+1}\right)\cos\left(x/2^{n+1}\right)\right)\tag{2b}\\ &=\frac1{x^3}\sum_{n=0}^\infty2^{n+1}\sin\left(x/2^{n+1}\right)\left(1-\cos\left(x/2^{n+1}\right)\right)\tag{2c}\\ &=\frac1{x^3}\sum_{n=0}^\infty\frac{2^{n+1}\sin^3\left(x/2^{n+1}\right)}{1+\cos\left(x/2^{n+1}\right)}\tag{2d}\\ &=\frac12\sum_{n=0}^\infty\frac1{2^{2n+2}}\left[(1-\epsilon)^3,\frac1{1-\epsilon/2}\right]\tag{2e}\\ &=\frac16\left[(1-\epsilon)^3,\frac1{1-\epsilon/2}\right]\tag{2f} \end{align} $$ Explanation:
$\text{(2a)}$: apply $\lim\limits_{x\to0}\frac{\sin(x)}x=1$ and the telescoping sum
$\text{(2b)}$: $\sin(2x)=2\sin(x)\cos(x)$
$\text{(2c)}$: factor
$\text{(2d)}$: $\sin^2(x)=(1-\cos(x))(1+\cos(x))$
$\text{(2e)}$: apply $(1)$ with $[a,b]$ representing a number in that range
$\text{(2f)}$: evaluate the sum

Therefore, by $(2)$ and the Squeeze Theorem, $$ \lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag3 $$

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The Approximations

The limit $(3)$ gives the following approximation $$ \sin(x)\approx x-\frac{x^3}6\tag4 $$ Since the Binomial Theorem gives $(1-x)^2=1-2x+x^2\approx1-2x$, we have $\sqrt{1-x}\approx1-\frac x2$. $$ \begin{align} \cos(x) &=\sqrt{1-\sin^2(x)}\tag{5a}\\ &\approx\sqrt{1-x^2}\tag{5b}\\ &\approx1-\frac{x^2}2\tag{5c} \end{align} $$ Finally, since the Geometric Series gives $\frac1{1-x}=1+x+x^2+\dots\approx1+x$, $$ \begin{align} \tan(x) &=\frac{\sin(x)}{\cos(x)}\tag{6a}\\ &\approx\left(x-\frac{x^3}6\right)\left(1+\frac{x^2}2\right)\tag{6b}\\ &\approx x+\frac{x^3}3\tag{6c} \end{align} $$