How to justify small angle approximation for cosine

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

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

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 $\cos(\theta) \approx 1-\theta^2/2$.