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Given $a$ rational, find all $\alpha$ such that the expression

$$\frac{\cos \alpha - a \sin \alpha}{a \cos \alpha + \sin \alpha}$$ is rational

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  • $\begingroup$ What have you tried? What progress have you made? $\endgroup$
    – saulspatz
    May 20 '21 at 17:35
  • $\begingroup$ @saulspatz I only know that $a = \frac{1}{5}$ and $\alpha = \pi/4 rad $ works. It gives you $2/3$. $\endgroup$ May 20 '21 at 17:38
  • $\begingroup$ Doesn't $\alpha=\frac\pi4$ work for any rational $a\neq-1$? $\endgroup$
    – saulspatz
    May 20 '21 at 17:41
  • $\begingroup$ @saulspatz The idea is to vary $\alpha$ and $a$ at the same time. If something is given, I prefer to set $a$, and then search $\alpha$. $\endgroup$ May 20 '21 at 17:43
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    $\begingroup$ Hint: $$\frac{a^{-1} - \tan\alpha}{1 + a^{-1}\tan\alpha}=\tan(\alpha-\arctan a^{-1}).$$ $\endgroup$
    – user65203
    May 20 '21 at 18:15
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If $\cos\alpha=0$ then the expression equals $-a$ and so is rational. Suppose therefore, that $\cos\alpha\neq0$. On dividing numerator and denominator by $\cos\alpha$, the expression becomes $$\frac{1-a\tan\alpha}{a+\tan\alpha}\tag1$$ which is clearly rational if $\tan\alpha$ is rational.

On the other hand, setting $(1)$ equal to $\frac pq$ and solving for $\tan\alpha$ gives a rational expression for $\tan\alpha$, so that we see that $(1)$ is rational if and only if $\tan\alpha$ is rational.

If you want values such that $\frac\alpha\pi$ is rational, then Niven's Theorem says that this only occurs when $\tan\alpha\in\{-1,0,1\}$

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    $\begingroup$ divide by cos will not give your expression, it will give $\frac{1-atan(\alpha)}{a+tan(\alpha)}$ $\endgroup$
    – Lac
    May 20 '21 at 18:43
  • $\begingroup$ @LSS Yes, you're right. It says that on my scratch paper, but I'm a terrible typist. Thanks, I've corrected it. $\endgroup$
    – saulspatz
    May 20 '21 at 18:45
  • $\begingroup$ If $\tan \alpha$ is rational then the expression is rational, But the other way implication is true? $\endgroup$ May 21 '21 at 15:49
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    $\begingroup$ @somenxavier If we set $(1)$ equal to a rational $r$ and clear denominators, we get a linear equation with rational coefficients for $\tan\alpha$, so the solution is rational. Is there a problem I'm missing? $\endgroup$
    – saulspatz
    May 21 '21 at 15:55
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Let $\beta = arctan(a)$

Then the expression becomes

$$\frac{\cos \alpha - \tan \beta \sin \alpha}{\tan \beta \cos \alpha + \sin \alpha}$$

Divide both the numerator and denominator by $\cos \alpha$ to get

$$\frac{1 - \tan \beta \tan \alpha}{\tan \beta + \tan \alpha} = \frac{1}{\tan (\alpha + \beta)}$$

So we want to find all $\alpha$ such that $\tan (\alpha + \beta)$ is rational.

So the set of all $\alpha$ is given by $arctan(Q)-\beta+\pi k=arctan(Q)-arctan(a)+\pi k$ where Q is a rational number and k is an integer.

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