A trigonometric coincidence

“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 $\sin \left(\frac{1}{3}\pi \right)$, 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 $2\times 5÷100=0.1$.

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!

Footnotes:

1. 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.