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?
$\endgroup$ 33 Answers
$\begingroup$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
$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$$
$\endgroup$ 1 $\begingroup$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 $\cos(\theta)\ge1-\dfrac{\theta^2}{2}$.
$\endgroup$ $\begingroup$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 $\cos(\theta) \approx 1-\theta^2/2$.
$\endgroup$
