Without computing it, conclude that $\tan(\frac{\pi}{8})$ is irrational. Let $ \theta\in (0,\frac{\pi}{4})$.
Prove that
$$ \sqrt{\sin(4\theta)}\in \Bbb Q \implies \tan(\theta)\notin \Bbb Q$$
and conclude that $\tan(\frac{\pi}{8})\notin \Bbb Q$.
I tried the contrapositve, but it became more difficult.
Any help will be appreciated.
Thanks in advance.
 A: Assume that $\tan\theta$ is rational with $0<\theta<\pi/4$, and that $\sin 4\theta$ is the square of a rational number. We're aiming for a contradiction.
Let $t=\tan\theta$. Then the complex number $1+it$ has argument $\theta$, and so $(1+it)^4$ has argument $4\theta$. Thus,
$$ \sin 4\theta = \frac{\mathop{\mathrm{Im}}\,(1+it)^4}{|(1+it)^4|} = \frac{4t(1-t^2)}{(1+t^2)^2}$$
Now suppose $t=p/q$ in lowest terms. Plugging in, we find
$$ \sin 4\theta = \frac{4p(1-p^2/q^2)}{q(1+p^2/q^2)^2} = \Bigl(\frac{2}{q^2(1+p^2/q^2)}\Bigr)^2 pq(q^2-p^2) $$
Thus, $pq(q^2-p^2)$ needs to be a perfect square.
However, since $p$ and $q$ are assumed coprime, no prime factor in $p$ appears in either $q$ or $q^2-p^2$, so $p$ itself has to be a square. Similarly $q$ has to be a square. Thus the third factor $q^2-p^2$ has to be square too.
So we have $p=x^2$, $q=z^2$, where $x$ and $z$ need to belong to a solution of
$$ x^4 + y^2 = z^4 $$
But Fermat's right triangle theorem tells us that this has no nontrivial solutions. (The trivial solution $y=0$ corresponds to $p=q$ and thus $\theta=\pi/4$, and $x=0$ corresponds to $\theta=0$, which are both out of the explicitly assumed range).
A: The typical proof of something like this is sometimes called "infinite descent" - you show that any solution would imply a smaller solution, which would imply a smaller solution... Since you're working with positive integers, there's only so far you could descend, so you get a contradiction.
So assume $\sqrt{3}=\frac{p}{q}$ where $p$ and $q$ have no common factors. Then we have $3=\frac{p^2}{q^2}$ or $3p^{2}=q^{2}$. But 3 is prime and divides the left hand side. So it also divides the right hand side. Since 3 divides $q^{2}$ it has to divide $q$. Say, $q=3q'$. Then we have $3p^{2}=9(q')^{2]}$. Dividing both sides by 3, we now have $p^{2}=3(q')^{2}$, so by the same argument $p = 3p'$  for some integer $p'$. This gives us a smaller pair of integers $p', q'$ such that $\sqrt{3}=\frac{p'}{q'}$, which gives us the start of our infinite descent. (Or, more simply, we've contradicted the initial premise that $p$ and $q$ have no common factor.)
Note this says $\sqrt{n}$ is irrational for any positive prime number $n$.
