Ask Uncle Colin: Rational Trigonometric Values

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?

Asking for a friend.

— 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 operations1 : 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.


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. as long as you don’t try to divide by zero []


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