# Morrie's Law

I suppose this ought to be a Dictionary of Mathematical Eponymy post, but it isn’t. So there ((for now, at least)).

The legend is that Richard Feynman, as a child ((History does not record whether he was a jerk at that point.)), was discussing trigonometry with his friend Morrie Jacobs in the Jacobs family leatherworks, when Morrie divulged that $\cos(20^o) \cos(40^o) \cos(80^o) = \frac{1}{8}$. This surprised and delighted Feynman, who eventually referred to it as Morrie’s law.

### Why does it work?

It works for very neat trigonometric reasons, related to the fact that $\sin(2x) = 2\sin(x)\cos(x)$.

In particular, $\sin(40^o)\sin(80^o)\sin(160^o) = \left(2\sin(20^o)\cos(20^o)\right)\times$ $\left(2\sin(40^o)\cos(40^o)\right)\times$ $\left(2 \sin(80^o)\cos(80^o)\right)$.

Most of those sines cancel!

We’re left with $\sin(160^o) = 8\sin(20^o)\cos(20^o)\cos(40^o)\cos(80^o)$.

However, $\sin(160^o) = \sin(20^o)$, so *those* cancel, as well! We’re left with $1 = 8\cos(20^o)\cos(40^o)\cos(80^o)$, which is Morrie’s law.

### Can it be extended?

Of course it can! You can do generalise the idea to $2^{n+1}\cos(x)\cos(2x)\cos(4x)\dots\cos\left(2^nx\right) = \frac{\sin(2^{n+1} x)}{\sin(x)}$. If you pick your value of $x$ so that the right hand side takes on specific values (for example, 1 is a handy one; I imagine one could contrive various fractions and square roots with a bit of thinking), you get an identity as delightful as Morrie’s.

Let’s try one that goes one step further: $16\cos(x)\cos(2x)\cos(4x)\cos(8x) = \frac{\sin(16x)}{\sin(x)}$.

Now I want $\sin(16x)$ and $\sin(x)$ to be the same - I could pick $\frac{180^o}{17}$, which is of course a lovely number. Isn’t it great that 360 has so many factors? It makes trigonometry so much simpler. In fact, I believe Morrie’s law is the furthest this can be taken with integer-degree angles ((Of course, in radians, it all works so much more nicely.))