Julian's musings

At a conference run by the BBO Maths Hub today, Jo Morgan mentioned that small angle approximations are a topic recently (re)introduced to the single maths A-level course, and many teachers may be unfamiliar with it.

During the day and on my journey home, I thought about this and some of the connections between it and other areas of the syllabus. So here are a few quick thoughts on ways we could think about them, making connections between this and other areas of the syllabus. I hope that this post offers some different perspectives on the topic.

This is a diagram probably familiar from most A-level textbooks (I don’t have one to hand, unfortunately). We have our familiar unit circle, and draw a right-angled triangle with angle $\theta$, opposite $\sin\theta$ and adjacent $\cos\theta$. We also see that the arc length subtended by the angle $\theta$ is $r\theta=\theta$ as the radius is 1. (We must be working in radians for this to be correct!) Already in this diagram, $\sin\theta$ and $\theta$ do not look very different, so $\sin\theta\approx\theta$. On the other hand, $\cos\theta$ looks pretty close to $1$, so we have $\cos\theta\approx1$. Visually, say using GeoGebra, we see that these approximations get better as $\theta$ gets smaller: the arc and the half-chord become closer and closer to each other. We can then work out $\tan\theta=\dfrac{\sin\theta}{\cos\theta}\approx {\theta}{1}=\theta$.

Another way of seeing this approximation to $\tan\theta$ is to draw the triangle with adjacent equal to $1$:

If we take $\sin\theta\approx\theta$, then we can work out a better approximation for $\cos\theta$ using the binomial theorem. We have, for small $\theta$ (positive or negative):

\[\begin{align*} \cos\theta &= \sqrt{1-\sin^2\theta} \\ &\approx \sqrt{1-\theta^2} \\ &= (1-\theta^2)^{\frac{1}{2}} \\ &= 1 - \tfrac{1}{2}\theta^2 + \cdots \end{align*}\]

where we have used the first two terms of the binomial expansion on the last line. So $\cos\theta\approx 1-\frac{1}{2}\theta^2$.

Another way of obtaining the approximation for $\cos\theta$ is to relate cos and sin using a double-angle formula:

\[\cos 2\theta = 1 - 2\sin^2\theta\]

so

\[\begin{align*} \cos\theta &= 1 - 2 \sin^2 \tfrac{1}{2}\theta \\ &\approx 1 - 2 \bigl(\tfrac{1}{2}\theta\bigr)^2 \\ &= 1 - \tfrac{1}{2}\theta^2 \end{align*}\]

where we have used $\sin\tfrac{1}{2}\theta\approx\tfrac{1}{2}\theta$ on the second line.

The approximations for $\sin\theta$ and $\tan\theta$ are also closely related to the shape of their graphs near the origin (though there is potentially some circular reasoning here - no pun intended!):

We have drawn the graphs of $y=x$ (red), $y=\sin x$ (green) and $y=\tan x$ (blue). Near the origin, the three graphs look very similar, so for small $x$, $\sin x\approx x \approx \tan x$.

This also tells us that at the origin, $\frac{d}{dx}(\sin x)$ and $\frac{d}{dx}(\tan x)$ equal $1$.

We can also argue in the opposite direction. If we have already convinced ourselves why the derivative of $\sin x$ is $\cos x$ using a different approach (for example, by using Rotating derivatives), then we can say that for small values of $x$, the graph of $y=\sin x$ is approximated by the tangent to the graph at $x=0$ (see A tangent is… for more on this point). We can calculate the tangent: since $\frac{d}{dx}(\sin x)=\cos x$ giving $\cos 0 = 1$, and $\sin 0 = 0$, the tangent has equation $y=x$. So for small $x$, $\sin x\approx x$.