Is this proof that $\tan a$ is transcendental correct?

Question

First, a proof that $\sin a$ is transcendental, where $a$ is algebraic not zero. Given that,

$$e^{ia}=\cos a + i\sin a$$

if $\sin a$ were algebraic, then $\cos a = \pm \sqrt{1-\sin^2a}\:$ is also algebraic, implying that

$$e^{ia}=\pm \sqrt{1-\sin^2a} + i\sin a$$

is algebraic, but $e^{ia}$ is transcendental (Lindemann theorem). Doing the same thing we can prove that $\cos a$ is transcendental.

If

$$\tan a = \frac{\sin a}{\cos a}$$

were algebraic, then $\sin a / \cos a$ is algebraic, implying that $\sin a$ and $\cos a$ are both algebraic, but this is false.

EDIT: Thanks! I get it now.

Answer 1

It is entirely possible for the quotient of transcendental numbers to be algebraic. Rather than that, you are looking for a relation like $$1+\tan^2 x=\frac1{\cos^2 x}$$

Answer 2

It might be clearer if you said something like

$$\sin \alpha = \pm\sqrt{\frac{1}{1+\frac{1}{\tan^2 \alpha}}}$$

making it clearer that if $\tan \alpha$ is algebraic then so too is $\sin \alpha$

Answer 3

From $$ \tan a = \frac{\sin a}{\cos a} $$ we get $$ \tan a = \frac{e^{ia} - e^{-ia}}{i(e^{ia}+e^{-ia})} $$ Solve this for $e^{ia}$. We conclude that $e^{ia}$ is the solution of a quadratic equation with algebraic coefficients.