Everyone knows the picture that explains instantly the small angle approximation to the sine function (as defined by the parametrisation of the unit circle): "what's the length of that arc?" "See how for small angles, it forms the opposite side of a triangle...?"

Cosine is more problematic; the corresponding annotation on Wikipedia to the diagram mentioned above reads:

H and A are almost the same length, meaning $\cos(\theta)$ is close to $1$ and $\frac{\theta^2}{2}$ [?!] helps to trim the red away [?!].

For the syllabus I teach, students must be able to differentiate sine and cosine from first principles using the above approximations. And certainly they don't need to understand the approximations; but it would be nice, wouldn't it...

Now everyone also knows that the small angle approximation for $\cos$ is just the truncated ($O(\theta^3)$) Taylor series, and it's fairly easy to see that for small $\theta$:

$$\cos(\theta)= \sqrt{1-\sin^2(\theta)} \approx \sqrt{1- \theta^2}$$

which $\approx 1- \frac{\theta^2}{2}$ by the binomial expansion for $\sqrt{1-x}$

...But my students don't know Taylor series or binomial expansions.

Question: Can one do any better?

  • $\begingroup$ And, no, drawing the graphs doesn't count. $\endgroup$ Feb 25 '12 at 23:34
  • $\begingroup$ How about completing the square and ignoring the fourth-order term in $\theta$? $\sqrt{1 - \theta^2} = \sqrt{1 - \theta^2 + \theta^4/4 - \theta^4/4} \approx \sqrt{\big(1 - \theta^2/2\big)^2}$. $\endgroup$
    – user856
    Feb 25 '12 at 23:40
  • $\begingroup$ @Tom Boardman: Does this count? $(1-\frac{x^2}{2})^2 =1-x^2+\frac{x^4}{4}\approx 1-x^2$ and therefore $\dots$. $\endgroup$ Feb 25 '12 at 23:43

You can use the double angle formula:

$$1 - \cos 2\theta = 2 \sin^2 \theta \sim 2\theta^2$$ and so

$$ \cos \theta \sim 1 - \frac{\theta^2}{2}$$

or ask them to prove that

$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$$

  • 1
    $\begingroup$ All of the answers here are great, but they've just learned the double angle formulae, so I'm accepting this one, as it's the one I'm probably going to teach. $\endgroup$ Feb 26 '12 at 11:02

One way to avoid the binomial expansion, is to note that for small $x$, $$ 1-\sqrt{1-x}=\left(1-\sqrt{1-x}\right)\frac{1+\sqrt{1-x}}{1+\sqrt{1-x}}=\frac{1-(1-x)}{1+\sqrt{1-x}}=\frac{x}{1+\sqrt{1-x}}\approx\frac{x}{2} $$ Therefore, $$ \sqrt{1-x}\approx1-\frac{x}{2} $$ Thus, for small $\theta$, $\cos(\theta)=\sqrt{1-\sin^2(\theta)}\approx1-\dfrac{\sin^2(\theta)}{2}\approx1-\dfrac{\theta^2}{2}$.

To finish things off, you can use that $\displaystyle\lim_{\theta\to0}\frac{\sin(\theta)}{\theta}=1$.

Post Script: It has been asked whether this is an over- or under-estimate.

For $x\ge0$, $\sqrt{1-x}\le1$, so we have $$ 1-\sqrt{1-x}=\frac{x}{1+\sqrt{1-x}}\ge\frac{x}{2} $$ Therefore, $$ \sqrt{1-x}\le1-\frac{x}{2} $$ Furthermore, $\sin(\theta)\le\theta$.

Thus, for small $\theta$, $\cos(\theta)=\sqrt{1-\sin^2(\theta)}\le1-\dfrac{\sin^2(\theta)}{2}\ge1-\dfrac{\theta^2}{2}$. This makes it difficult to determine that $\cos(\theta)\ge1-\dfrac{\theta^2}{2}$.


Do your students know the addition theorems? $\cos(\theta) = \cos^2(\theta/2)-\sin^2(\theta/2) = 1 - 2 \sin^2(\theta/2)$. Now if $\sin(\theta/2) \approx \theta/2$, you get immediately $\cos(\theta) \approx 1-\theta^2/2$.


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