Is there a number such that x is rational, and cos(x) is rational? Since I doubt this is true, is there a number such that x is algebraic and cos(x) is algebraic?

x=0?

Sorry for the above post. :-)

I think the less trivial answer is probably no. I seem to remember that there is a theorem that says something like if x is rational and nonzero then e^x is transcendental. If this extends further and we can say that if x is algebraic and nonzero then e^x is transcendental then we're done. Because, if x was algebraic and cos(x) algebraic then also sin(x) is algebraic because sin² (x)+cos² (x)-1=0 and so also is i.sin(x) and i.x. So we have that cos(x)+i.sin(x) is algebraic, i.e. e^ix is algebraic. But this would be a contradiction.

So, does anyone know if the theorem I used is true or not?

Yes, it's a special case of Lindemann's theorem. Lindemann's theorem says that if a₁, ..., a_n are distinct algebraic numbers and if b₁, ..., b_n are algebraic then:

b₁ e^a₁+¼+b_n e^a_n=0

implies that b₁=¼ = b_n=0.

I don't get it why not x=0??

love arun

Arun, x=0 is fine. The point is that x=0 is the only such number. What we've proved above is that there are no other x apart from that obvious one.

Oh!i get it...

love arun