0

I made a proof using isometry, but I'm not sure. Here is the proof:

Let A be a 2×2 matrix whose first row is equal to (cos(1), -sin(1)) and second row is equal to (sin(1), cos(1)). And define the sequence {X(n)} by

X(1)=(cos(1), sin(1)), X(n+1)=A^nX(1).

Since A is an isometry, norm(X(n+1)-X(n))=norm(X(2)-X(1)) =/0 for all natural numbers n. Hence {X(n)} is not a Cauchy sequnce.

One of {cos(n)}, {sin(n)} diverges because {x(n)} converges iff {cos(n)} and {sin(n)} both converges. Suppose f(n) be the one that diverges. Then sqrt[1-(f(n)^2] also diverges. This sequnce is equal to the other one. Hence, {cos(n)}, {sin(n)} both diverge.

JJG
  • 1
  • 1
    One issue is that $\cos(x)$ can as well be $-\sqrt{1-\sin^2(x)}$. Also, where did you show that it is impossible that both sequences converge ? – Peter Jan 04 '22 at 14:50
  • Thank you for your comment. I thought these two statemets below prove that cos(n) and sin(n) diverge. :
    1. X(n)=(a(n), b(n)) converges to (p, q) if and only if a(n) converges to p and b(n) converges to q. 2) If one of cos(n) and sin(n) diverges, the other one also diverges.
     – JJG Jan 04 '22 at 15:18
  • That $\sin(n)$ and $\cos(n)$ do not converge is a deep result. We can even prove that the values of each of the sequences are dense in $[-1,1]$. But I do not think that there is an easy way to show that. – Peter Jan 04 '22 at 16:38

1 Answers1

1

Let's prove $\{ \sin(n) \}$ diverges by way of contradiction.

Suppose that $\lim{\sin n}=a$. We know the equality: $$ \sin(n+1) - \sin (n-1) = 2 \cos(n) \sin (1) $$ taking limit $n \to \infty$ from both sides we obtain: $$ a-a=0=2 \sin(1) \lim_{n\to \infty}{\cos(n)} $$ thus $\lim_{n\to\infty}{\cos(n)}=0$. Again, we have: $$ \sin(2n)=2 \sin(n) \cos(n) $$ taking limit: $$ a =2 a \cdot 0 = 0 $$

But $\sin^2n+\cos^2n=1$, taking limit: $$ 0+0=1 $$ which is a contradiction.

Jamie Kamsky
  • 115