$\tan{\alpha} \approx \alpha$ if $\alpha$ is small

Question

I saw this a lot in physics textbook but today I am curious about it and want to know if anyone can show me a formal mathematical proof of this statement? Thanks!

Answer 1

Consider $\lim_{x\to 0} \frac{\tan x}{x}$ and apply the L'Hôpital's rule.

You'll get a limit of $1$ as $x \to 0$, proving the statement (relative error).

You can even get a bound on the error term with Taylor's expansion of $\tan x$, which is $\mathcal{O}(x^3)$ (absolute error).

Answer 2

The one-sentence answer is that the Taylor series for tangent at zero is $x + O(x^3)$. So it is actually quite a good approximation.

Answer 3

This is the most basic way:

$$\lim_{\alpha \to 0} \frac{\tan \alpha}{\alpha} = \lim_{\alpha \to 0} \frac{\sin \alpha}{(\cos \alpha)\alpha} = \lim_{\alpha \to 0} \frac{1}{\cos\alpha}\lim_{\alpha \to 0} \frac{\sin \alpha}{\alpha} = \frac{1}{\cos(0)}\lim_{\alpha \to 0} \frac{\sin \alpha}{\alpha}=\lim_{\alpha \to 0}\frac{\sin \alpha}{\alpha}=1.$$

The last limit follows from a simple geometric argument and the Squeeze Theorem. You don't even need to use derivatives. The Squeeze Theorem is in any good calculus textbook. The geometric argument has appeared here: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?, and at least the first answer there has a nice diagram (indeed, your question is almost a duplicate of that question). You can use L'Hopital's Rule, and it works, but L'Hopital's rule depends on being able to differentiate trig functions, and the derivatives of the trig functions are proven using the last limit above, which is what you're trying to prove!

Some of the other answers are more informative in that they tell you how accurate the approximation $\tan(\alpha) \approx \alpha$ is for $\alpha \approx 0$ (mine doesn't). But most or all of them use some tool such as a Taylor series or L'Hopital's Rule, which requires you to know derivative(s) of $\tan$ and/or $\sin$. Those derivatives are computed using the last limit which appears above. So if you're using L'Hopital's Rule, for example, you're probably implicitly using the limit $\sin(\alpha)/\alpha \to 1$ as $\alpha \to 0$, which is almost exactly the same as what you're trying to prove (you have to use $\cos(0) =1$, of course) Which kind of answer you prefer is up to you.

Answer 4

As already expressed in another answer, this can be formulated in a formal way by taking the Taylor expansion of $\tan\alpha$ around $0$ to get: $$\tan\alpha = \alpha + \frac{1}{3}\alpha^3 + \ldots = \alpha + O(\alpha^3)$$ where $O(\alpha^3)$ denotes some function that goes to zero approximately at the same rate as $\alpha^3$ does. Formally, a function $f(x)$ is $O(g(x))$ if $\limsup_{x\rightarrow 0}\left|\frac{f(x)}{g(x)}\right|<\infty$.

In general, looking at the Taylor expansion of a function around some point is a very good way to get an approximation of the function near that point. For example we have: $$\sin(x) = x+O(x^3)$$ $$\cos(x)=1+O(x^2)$$ $$e^x=1+x+\frac{1}{2}x^2+O(x^3)$$ around zero. This is very useful in various areas of mathematics and physics.

Answer 5

The formal proof would involve demonstrating that, for any value $\epsilon<1$, one can find a value $\omega>0$ such that

$$ \forall \alpha\in(-\omega,\omega), |\tan(\alpha)-\alpha|<\epsilon |\alpha| $$

That is, for all values of $\alpha$ between $-\omega$ and $\omega$, the difference between $\tan(\alpha)$ and $\alpha$ is smaller than $\epsilon |\alpha|$. The formal logical expression actually takes the form

$$ \forall \epsilon<1, \exists \omega>0 \text{ s.t. }\forall \alpha\in(-\omega,\omega), |\tan(\alpha)-\alpha|<\epsilon |\alpha| $$

It isn't trivial to prove it formally, requiring some proper analysis. The informal proofs provided by Eric Auld and ftfish should be sufficient to demonstrate the it in a less rigorous manner.