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In a similar vein to this question, I am trying to understand the occurrence of rational solutions $\lambda$ to the following equation $$\cos(\lambda \pi) = \cos^2 (a\pi) - \cos ( b\pi ) \sin^2 \left( a\pi \right),$$ where above $a,b \in \mathbb{Q}$.

In particular, I want to conjecture that if $a$ is not an integer multiple of $\frac{1}{4}$, and if $b$ is not an integer multiple of $\frac{1}{2}$, then $\lambda$ must be irrational. These two constraints are simply coming from the fact that $\cos(\lambda \pi) = \cos^2(a \pi)$ has no rational solutions $\lambda$ if $a$ is not an integer multiple of $\frac{1}{4}$ (see linked question above), and the observation that if $b = 1$ then $$\cos^2 (a\pi) - \cos ( b\pi ) \sin^2 \left( a\pi \right) = 1$$ and if $b = 2$ then $$\cos^2 (a\pi) - \cos ( b\pi ) \sin^2 \left( a\pi \right) = \cos^2 (a\pi) - \sin^2 \left( a\pi \right) = \cos(2a \pi).$$

I have tried to mimic the approach used in the linked question above, but have been unable to make progress using that technique or others. I wondered if I am missing something and there is a clear way to show this, or if perhaps my constraints are not tight enough and the answer is more complicated?

Thanks!

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  • $\begingroup$ May I ask, why are you interested in this specific equation? $\endgroup$
    – Fedor Petrov
    Commented Feb 1 at 3:45
  • $\begingroup$ A useful trick is to take the average over all algebraic conjugates, using the fact that the sum of primitive roots of 1 of degree $k$ equals $\mu(k)$ $\endgroup$
    – Fedor Petrov
    Commented Feb 1 at 3:48
  • $\begingroup$ Roughly, I came to this question in trying to prove the universality of a gate set in quantum computation. I essentially have a gate that performs a rotation by $\lambda \pi$ about a particular axis, such that the rotation is equal to the arccos of the LHS of the equation I propose in my question. Whenever lambda is irrational, then this rotation can be used to approximate any rotation about the axis, which is quite useful. $\endgroup$
    – Mary_Smith
    Commented Feb 1 at 4:07
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    $\begingroup$ note that $b=1/2, a=1/4, \lambda=1/3$ works so the correct guess would be $b$ not a half integer multiple and $a$ not a quarter integer multiple; a simple manipulation using $1=\cos^2 a\pi +\sin^2 a \pi $ and subtracting reduces the above to the equation $\sin \lambda_1 \pi=\sin a\pi \sin b_1\pi$ where $\lambda_1=\pm \lambda/2, b_1=b/2$ $\endgroup$
    – Conrad
    Commented Feb 1 at 5:00
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    $\begingroup$ In Conrad's reformulation it is addressed and completely solved by David Speyer here mathoverflow.net/questions/458827/… $\endgroup$
    – Fedor Petrov
    Commented Feb 1 at 6:26

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