Trigonometry – How to Justify Small Angle Approximation for Cosine

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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?

Best Answer

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

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