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 $\sin\left(\frac 13 \pi\right)$, 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!

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.

  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. []

Share

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

On twitter