# Find all $\alpha$ such that $\frac{\cos \alpha - a \sin \alpha}{a \cos \alpha + \sin \alpha}$ is rational (given $a$ rational)

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

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

• What have you tried? What progress have you made? May 20 '21 at 17:35
• @saulspatz I only know that $a = \frac{1}{5}$ and $\alpha = \pi/4 rad$ works. It gives you $2/3$. May 20 '21 at 17:38
• Doesn't $\alpha=\frac\pi4$ work for any rational $a\neq-1$? May 20 '21 at 17:41
• @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 $\alpha$. May 20 '21 at 17:43
• Hint: $$\frac{a^{-1} - \tan\alpha}{1 + a^{-1}\tan\alpha}=\tan(\alpha-\arctan a^{-1}).$$
– user65203
May 20 '21 at 18:15

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 $$\tan\alpha\in\{-1,0,1\}$$

• divide by cos will not give your expression, it will give $\frac{1-atan(\alpha)}{a+tan(\alpha)}$
– Lac
May 20 '21 at 18:43
• @LSS Yes, you're right. It says that on my scratch paper, but I'm a terrible typist. Thanks, I've corrected it. May 20 '21 at 18:45
• If $\tan \alpha$ is rational then the expression is rational, But the other way implication is true? May 21 '21 at 15:49
• @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? May 21 '21 at 15:55

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.