Elementary proof (involving no series expansion) that $\tan x \approx x+\frac{x^3}{3}$ for small $x$ For small values of $x$, the following widely used approximations follow immediately by Taylor expansion:

*

*$\sin x \approx x$

*$\cos x \approx 1-\frac{x^2}{2}$

*$\tan x \approx x +\frac{x^3}{3}$
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.

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.

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:
 
