I gave a talk (some months ago now) on the history of $\pi$ (which is well discussed in my unreliable history of maths, Cracking Mathematics, available wherever good books are sold.)

At one point, I put up a slide generally excoriating degrees as a measurement of angle, and stating that for small $x$, you have $\sin\br{x^\circ}\approx \frac{\pi}{180}x$.

And one member of the audience was Not Having It. He was *certain* that the $x$ on the right also needed a $^\circ$ after it.

### Here's why (I think) he is mistaken

Let's start with the idea of units. If I write something like 15m, the m means "this number represents a distance, in the proportion such that 40,000,000 of them make a circumference of the earth" (or whatever the precise definition is these days).

My understanding of the $^\circ$ symbol is that - even though angles are dimensionless - it functions as a unit: it states "this number represents an angle, in the proportion such that $360^\circ$ is a complete revolution."

The sine function is a mapping from *angles* to *real numbers* (in particular, real numbers between -1 and 1). It's usually convenient to deal with angles in *radians*, which are angles in the proportion such that $2\pi$ represents a complete revolution - and it's so convenient to deal with it this way that we don't normally mention the unit - it's implied. Angles (in radians) generally behave just like numbers, so there's no difficulty.

Putting a $^\circ$ on the right-hand side of the equation would state "the right hand side is an angle" - which it isn't - and that 360 of those angle-units made up a circle - which they don't.

It would perhaps have been better to write it as $\frac{\pi}{180^\circ}x^\circ$ - but I stand by what I stated.

And I can only apologise to the audience for what must have seemed like an interminable, technical conversation.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

## Andrew Old

I think it’s pretty obvious that you were right. Difficult to see any grounds for objecting that aren’t just a massive misunderstanding of basic notation.

## Colin

Thanks, Andrew.