Written by Colin+ in ninja maths, trigonometry.

Radians - as I've ranted before - are the most natural way to express angles and do trigonometry. No ifs, no buts, degrees are an inherently inferior measure and the sooner they're abolished, the better. (In other news, the campaign to replace the mishmash of units called 'time' by UNIX timestamps starts here.)

Now, one of the reasons radians rock is, it's very easy to estimate the sine of a small angle in radians: it's almost exactly the same as the angle. That makes trigonometry super-easy. For instance, $\sin(0.05)$ is very close to 0.05. In fact, it's 0.4998 - which isn't too shabby at all. Even up as far as $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$, the error is only about 11%. That's not quite so good, but by making clever use of some other approximations (which I'll leave for another article), you can get much closer.

Anyway. When you have a small angle - let's say up to 30º - you can get a pretty good estimate for its sine by simply converting to radians. You remember from before that a degree is about 7/400 of a radian? If you have 8º, you can say that that's 0.14 radians, so $\sin(8^\circ)$ is roughly 0.14. (It's 0.1392).

What's that? Oh, do keep up. To work out $8 \times \frac{7}{400}$, you can cancel a 4 to get $\frac{2 \times 7}{100}$, which is $\frac{14}{100}$ or 0.14.

Going the other way - even supposedly sensible exam boards are wont to ask for answers to trigonometry questions in degrees once in a while - isn't much harder: you divide by $\frac{7}{400}$ (or multiply by $\frac{400}{7}$). To get $\sin^{-1}(0.28)$, you'd say that's $\frac{28}{100} \times \frac{400}{7}$, and cross-cancel to get 16º. The answer is 16.28º.

You can also use this to figure out the cosine of angles close to a right angle using the identity $\cos(90-x)^\circ \equiv \sin(x)^\circ$. If you want $\cos(85^\circ)$, you can say "ah! that's the same as $\sin(5^\circ)$, so it's $\frac{35}{400}$, about 0.088. (It's 0.0872, which is annoyingly close).

Oh - and a couple more things. A special bonus: the tangent function behaves just like sine for small angles, because $\tan(x) \equiv \frac{\sin(x)}{\cos(x)}$ and $\cos(x)$ is close to 1 for small x - which means you can estimate $\tan$ and $tan^{-1}$ in much the same way. Finally, the identity from before means that for numbers near 90º, you can say $\tan(90-x)^\circ \equiv \frac{\sin(90-x)^\circ}{\cos(90-x)^\circ} = \frac{\cos(x)^\circ}{\sin(x)^\circ} = \frac{1}{\tan(x)^\circ}$. If you're interested in $\tan(89^\circ)$, you can say that's $\frac{1}{\tan(1^\circ)}$; $\tan(1^\circ)$ is about $\frac{7}{400}$, so $\tan(89^\circ)$ is roughly $\frac{400}{7} = 57.1$. It's actually 57.3.

Recognise that number? It's one radian converted to degrees. Why would that be?