If $x$ is rational and $\sin(x)$ is irrational, then is $\cos(x)$ irrational? I am wondering if the following implication is true. I know it is not true if $x$ is irrational, as clearly $x =\pi/3$ is a counterexample.
$$\text{$x$ is rational and $\sin(x)$ is irrational} \implies \cos(x) \text{ is irrational}$$
I know irrationals are not closed under multiplication or addition. However, this seems true.
If there is an obvious counterexample I am missing, I apologize.
 A: A form of the Lindemann-Weierstrass theorem states that if $a_1,...,a_n$ are algebraic numbers and $\alpha_1,...,\alpha_n$ are distinct algebraic numbers, then the only solution to :
$$
a_1e^{\alpha_1} + ... + a_n e^{\alpha_n} = 0
$$
is $a_i = 0$ for all $i$.

Suppose that for some $x$ non-zero rational (in fact, algebraic), we have $\sin x$ is rational. In particular, $\sin x = \frac{e^{ix} - e^{ix}}{2i}$, so we get $e^{ix} - e^{-ix} = 2iq$ for some rational $q$. Rearranging gives us :
$$
(1)e^{ix} + (-1)e^{-ix} + (-2iq)e^0 = 0
$$
If $x$ is non-zero , then the above contradicts the LW theorem because $ix,-ix$ and $0$ are distinct algebraic numbers. This shows that $\sin x$ is in fact irrational. But we can say more : see if you can prove that $\sin x$ is transcendental using this fact.

From here, a similar trick can be used to show that if $x$ is non-zero algebraic then $\cos x$ is irrational. Thus your implication is clear since both sides of it are always true unless $x=0$ in which case both sides are false, so the implication holds always.
