Written by Colin+ in core 3, trigonometry.

"Hm," I thought, "that's odd."

I don't often work in degrees, but the student's syllabus insisted. And $\sin(50º)$ came up. It's 0.7660, to four decimal places. But... I know that $\sin\left(\frac 13 \pi\right)$, er, sorry, $\sin(60º)$ is 0.8660 -- a difference that's pretty close to $\frac 1{10}$.

Which got me wondering: is that something interesting, or just a coincidence?

Well... it's a bit of both, I think.

There's a nice trick for finding the difference between (say) two sines, two cosines or one of each, and it's to use the formulas in your A-level book, the ones you never really look at, the ones after the compound angle formulas. 1

The one you're interested in is $\sin(A) - \sin(B) = 2 \cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$.

In this case, it gives $\sin(60º) - \sin(50º) = 2 \cos(55º) \sin(5º)$ which, naturally, comes to just under 0.1. But why should it?

$\cos(55º)$, if you ask your calculator, is about 0.5736. Is that a familiar number? Our straw poll said... heehaw. However, it looks a bit familiar to me -- 1 glorious radian is about 57.3 horrible degrees ($\frac {180}{\pi} \approx 57.296$ -- roughly $100 \cos(55º)$.

Meanwhile, in radians, $\sin(\theta) \approx \theta$, which means in degrees, $\sin(xº) \approx \frac{\pi x}{180}$.

That means, $2 \cos(55º) \sin(5º) \approx 2 \times \frac{180}{100\pi} \times \frac {5\pi}{180}$. Cancel the 180s and the $\pi$s, obviously, to get $2 \times 5 \div 100 = 0.1$.

An upshot of this is that two angles the same distance either side of 55º will (for small differences) work out to roughly $\frac{1}{100}$ of the difference between them in degrees.

There's another coincidental pair that I've found: $\sin(75º) - \sin(60º) \approx 0.1$ as well, although I can't see any particular justification for that. However, those are both values that can be found exactly:

$\sin(75º) = \sin(45º + 30º) = \sin(45º)\cos(30º) + \cos(45º)\sin(30º) = \frac{\sqrt 2}{2} \frac{1}{2} + \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} = \frac{\sqrt{2} + \sqrt {6}}{4}$.

$\sin(60º) = \frac{\sqrt{3}}{2}$

So the difference between them is $\frac{\sqrt{2} + \sqrt {6} - 2\sqrt{3}}{4} $, which means that $\sqrt{2} + \sqrt{6} - 2\sqrt{3} \approx 0.4$ -- in fact, it's 0.3996 to 4dp.

Last few: $\sin(75º) - \sin(50º) \approx 0.2$; $\sin(40º)-\sin(20º) \approx 0.3$; $\sin(35º)-\sin(10º)\approx 0.4$, and $\sin(80º)-\sin(29º) = 0.49998$, which is very close indeed to a half!

- Technically, they're fair game, although I'm pretty sure I've only ever seen them in Solomon papers -- with the recent trend towards more involved C3 questions, I wouldn't be at all surprised to see them this summer. [↩]