Dear Uncle Colin,

You know how sometimes $\sin(2x)$ is rational and $\sin(5x)$ is rational and $\sin(7x)$ is rational, right? Would that necessarily mean that $\sin(12x)$ is rational?

— Perhaps You THink All Geometry’s On Right Angled Stuff

Hi, PYTHAGORAS, I believe it does! (In fact, I can prove it.)

I’m going to use two identities for it:

• $\cos(2A) \equiv 1 - \sin^2(A)$
• $2\cos(B)\sin(C) \equiv \sin(B+C) - \sin(B-C)$

… as well as the fact that the rationals are closed under the four basic operations ((as long as you don’t try to divide by zero)) : if you add, subtract, multiply or divide two rational numbers, you get another rational number.

The first of those identities tells you that $\cos(10x)$ and $\cos(14x)$ are both rational, because $\sin(5x)$ and $\sin(7x)$ are.

The second tells you that $2\cos(14x)\sin(2x) \equiv \sin(16x) - \sin(12x)$, which is rational, and that $2\cos(10x)\sin(2x) \equiv \sin(12x) - \sin(8x)$, which is also rational.

If you subtract those, you get $\sin(16x) - 2\sin(12x) + \sin(8x)$, which has to be rational.

Then you can use the second identity in reverse to say $\sin(16x) + \sin(8x) \equiv 2\cos(4x)\sin(12x)$.

That means you can rewrite $\sin(16x) - 2\sin(12x) + \sin(8x)$ as $2\cos(4x)\sin(12x) - 2\sin(12x)$, or even $2\sin(12x) \left[ \cos(4x) - 1\right]$. That’s rational, and so is $\cos(4x) - 1$, so $\sin(12x)$ must be!

I hope your friend gets some comfort from this.

— Uncle Colin

* Edited 2016-08-28 to fix an HTML error.