Season 12 Episode 4
By viewer request, we're talking about why we use radians in trigonometry when we want to do calculus, with help from Oxford student Judah.
Further Reading
Series for Sine
In the livestream, I presented two series expressions as an alternative definition of sine and cosine.
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n \, x^{2n+1}}{(2n+1)!}$$
$$\cos x =1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n \, x^{2n}}{(2n)!}$$
How can we get from the series to $\sin^2 x+\cos^2 x=1$?
It’s tempting to take the series and square it. Don’t do that! That’s a real mess.
Instead, let’s use the fact, from those series, that the derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$. If we’re going to show that $\sin^2 x +\cos^2 x=1$ then perhaps we could first show that $\sin^2 x +\cos^2 x $ is constant. How do you show that something’s constant? Easy, you differentiate it and check that you get zero.
To differentiate $\sin^2 x+\cos^2 x$? I need the chain rule. This is the highly useful rule that the derivative of $f(g(x))$ with respect to $x$ is $f’(g(x))\times g’(x)$, where $f’$ denotes the derivative of $f$ and similarly $g’$ denotes the derivative of $g$.
The derivative of $\sin^2 x+\cos^2 x$ is $2\sin x (\cos x) + 2 \cos x (-\sin x)$. That’s zero! So the expression $\sin^2 x+\cos^2 x$ must be constant. What’s the value of that constant? Well let’s just substitute in our favourite value of $x$. Perhaps $x=0$ will be easiest to work with. That gives $0^2+1^2$. So we see that the value of the constant is 1. Therefore $\sin^2 x+\cos^2 x=1$ for all $x$.
From there, we can see that the point $(\cos x, \sin x)$ lies on the unit circle, and we’re close to recovering our classic intuition of an angle. We’ll get there via arc length. As the point $(\cos x, \sin x)$ moves around the unit circle, we can check that it moves at constant speed 1. This will match with our understanding that changing $x$ by some amount traces out a predictable amount of the circumference of the circle, which we can think of either as an arc length or as an angle; for radians, those are the same thing.
To find the speed of the point $(\cos x, \sin x)$ we take the absolute value of the derivative. I can’t remember when I learned to take the derivative of a vector where each component depends on some variable $x$, or whether I was disappointed to find out that you just differentiate each component. Anyway, your turn to learn now (perhaps!); to differentiate a vector expression like this, you just differentiate each component.
We get $(-\sin x, \cos x)$. Tangent fact: this is a vector representing the direction of the tangent. If you know about scalar product / dot product then you can check that this is orthogonal to the radius.
The magnitude of that vector (the speed, as opposed to the velocity) is just $\sqrt{\sin^2 x+\cos^2 x}$, which is 1. So as $x$ changes, this point traces out the unit circle at uniform speed. That’s what arc length measures. And then, because of how radians work, that’s the angle. So this exactly corresponds to our ideas that $(\cos x, \sin x)$ should move according to an "angle" at the origin that increases steadily with arc length.
Phrased like that, it might sound almost like I’m saying there’s no such thing as angle, like I’m pulling the rug out from under your feet. I want to reassure you that this is a normal way to feel when you start playing around with definitions and consequences; whenever there are two rival definitions it’s possible to get attached to one as more "natural", and balk at the idea of deducing it from something else. But perhaps this is a reminder that definitions are not natural, given to us on stone tablets from a deity. We get to choose the starting points.
Parsec
It came up in passing, so here’s the modern "official" definition of the parsec.
The International Astronomical Union in 2015 decided that a parsec is exactly $\frac{648 000}{\pi}$ au, where 1au is 1 astronomical unit, which is defined to be precisely 149 597 870 700 metres. No triangles mentioned!
That’s a strange fraction, isn’t it? It comes from an angle of one arcsecond, which is $\frac{1}{3600}^\circ$ (one sixtieth of one sixtieth of a degree, like how a second is one sixtieth of a sixtieth of an hour), but that's converted to radians, which would be an angle of $\frac{2\pi}{360\times3600}$ radians. Then it’s the reciprocal of that, because the diagram we’re imagining here is a massive circle with radius one parsec, that tiny angle marked at the centre between two radii, and an arc length of 1au around the outside. Still not a triangle, but very close to one!
Gradians and French Metric Time
During the French Revolution, there was an attempt by the French government to decimalise everything; this leads to ideas like "90 degrees in a right angle is not quite enough, let’s have 100 slightly smaller degrees, and call them gradians". You’ve maybe noticed the option on your calculator after degrees and radians (D/R/G). I think there was a point where I had noticed the option on my calculator, and I was hoping that this mysterious G unit would be better than radians by a similar margin to the amount that radians are better than degrees. Sadly not.
The French also tried to decimalise time; 10 metric hours in a day, 100 metric minutes in a metric hour, and 100 metric seconds in a metric minute. This was officially in use for 2 years, or 73,000,000 metric seconds.
The Tau Manifesto
We mentioned this in passing and I (James) was quite dismissive. Don’t @ me. You can read about tau at https://www.tauday.com/tau-manifesto. Here’s a Wolfram blog post from 2015 trying to work out if it’s actually a good idea 2 Pi or Not 2 Pi? | Wolfram Blog. Spoilers; it’s a bit of a dead heat, and then the author suggests that really we should have a name the constant \(2\pi i \) where \(i^2=-1\)...
Exponentials
The exponential function $e^x$ can also be written as a power series
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$
If you differentiate term-by-term, you can recover the excellent fact that the derivative of $e^{x}$ is $e^{x}$. From there, we can do things like solve differential equations and model rapidly increasing growth.
Even better, this gives us a definition of the number $e$. It’s just
$$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots$$
Hyperbolic Functions
Two more functions that I’d like to show you
$$\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdots = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$$
$$\cosh x =1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$$
These are like $\sin x$ and $\cos x$ for people who don’t like tracking minus signs. The derivative of $\sinh x$ is $\cosh x$ and the derivative of $\cosh x$ is $\sinh x$. No minus signs to remember there!
They’re normally introduced in terms of exponentials as
$$ \sinh x = \frac{e^x-e^{-x}}{2}\quad \text{and}\quad \cosh x = \frac{e^x+e^{-x}}{2}$$
With that observation, it’s not too hard to check that they satisfy weird versions of the standard trigonometric relationships, with facts like $\cosh^2 x - \sinh^2 x = 1$ (I lied, there are some minus signs to remember).
Hyperbolic functions can be used to describe "Lorentz boosts" in Special Relativity, and people even sometimes refer to the concept of "hyperbolic rotation" by the "hyperbolic angle", which I have personally never been able to visualise, even slightly!
Knights and Knaves
This was a passing mention in the episode, but I can’t resist linking you to several classic puzzles involving "knights" (characters who always tell the truth) and "knaves" (characters who always lie). In these puzzles, you’re supposed to imagine that all the characters have perfect knowledge, are excellent at logical deduction, and always follow the rules. If you make these assumptions about people you meet in real life, then your mileage may vary.
Example: You meet two characters, X and Y, known to be either knights or knaves. X says: "At least one of us is a knave." What are X and Y?
Solution: If they’re both knights, then the statement X makes wouldn’t be true, which wouldn’t make sense (X tells the truth in that case!). So at least one of them is a knave; and therefore the statement is true. So X is telling the truth and is therefore a knight. That leaves Y as a knave, because at least one of them is.
These puzzles were popularised by Raymond Smullyan, whose books "What is the Name of This Book?" and "The Lady or the Tiger?" have hundreds of examples.
MAT 2010 Q3
Past MAT papers are available on the departmental website. Note that Oxford Mathematics now uses the TMUA for admissions but past MAT questions may still be an interesting source of challenging mathematics for you.
The question mentioned on OOMC was MAT 2010 Q3.
If you want to get in touch with us about any of the mathematics in the video or the further reading, feel free to email us on oomc [at] maths.ox.ac.uk.