# How to justify small angle approximation for cosine

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?

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

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

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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. –  Tom Boardman 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 $\cos(\theta)\ge1-\dfrac{\theta^2}{2}$.

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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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