Rational & Irrational

Since $\cos2\theta = 2\cos^2\theta - 1$ , we must have $\cos^2\theta \in \mathbb{Q}$ if we want $n \ge 2$ . Since $\cos3\theta \; = \; 4\cos^3\theta - 3\cos\theta \; = \; (4\cos^2\theta - 3)\cos\theta$ we must have $\cos^2\theta = \tfrac34$ if we want $n \ge 3$ .

• If $\cos\theta = \tfrac12\sqrt{3}$ then $\theta \equiv \pm \tfrac16\pi \pmod{2\pi}$ , and so $\cos2\theta = \tfrac12$ , $\cos3\theta = 0$ , $\cos4\theta = -\tfrac12$ , but $\cos5\theta = -\tfrac12\sqrt{3}$ .

• If $\cos\theta = -\tfrac12\sqrt{3}$ then $\theta \equiv \pm\tfrac56\pi \pmod{2\pi}$ , and so $\cos2\theta = \tfrac12$ , $\cos3\theta = 0$ , $\cos4\theta = -\tfrac12$ and $\cos5\theta = \tfrac12\sqrt{3}$ .

Thus $n = \boxed{4}$ .

We can solve the problem in two parts. First we show that we can meet the conditions when n=4. As an example take $x=\frac{\pi}{6}$ . Then

$cos(x)=cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\notin Q$ as required. And for n=2,3,4 the other conditions are met

$cos(2x)=cos(\frac{2\pi}{6})=\frac{1}{2}\in Q$

$cos(3x)=cos(\frac{3\pi}{6})=0\in Q$

$cos(4x)=cos(\frac{4\pi}{6})=-\frac{1}{2}\in Q$

Next we show that the conditions always fail when n=5. Using the well known addition formula it is not too hard to show that

$cos5x=2cos(4x)cos(x)-cos(3x) \dots(1)$

Now suppose we have managed to choose an x so that

$cos(x)\notin Q$

$cos(3x)\in Q$

$cos(4x)\in Q$

Then equation (1) shows that $cos(5x)=\text{rational}\times \text{irrational}-\text{rational}=\text{irrational}\notin Q$

And so the maximum value of n is $\boxed{\text{4}}$

'Tetchy bits and bobs'

One thing that could go wrong with this argument is that in (1) we might have $cos(4x)=0$ which would eliminate the irrational part of the RHS. However in this case we have $x=\frac{\pi}{8}$ in which case $cos(3x)=cos(\frac{3\pi}{8})=\frac{\sqrt{2-\sqrt{2}}}{2}\notin Q$ and we have already failed!