Question

QUESTION 10 The sequence sin(1), sin(2), sin(3), sin(4), ..... O has a constant subsequence in (-1,1] O has no convergent subsequence O converges O has a convergent subsequence in (-1,1]

Answer 1

We know that $-1 \leq \sin (x) \leq 1$ for all real number x .

So the sequence $\sin(1) , \sin (2) , \sin (3) , \sin (4) ,....$ is a bounded sequence of real number.  

By Bolzano-weierstrass theorem every bounded sequence have a convergent subsequence so the sequence $\sin(1) , \sin (2) , \sin (3) , \sin (4) ,....$ has a convergent subsequence.  

Answer : has a convergent subsequence in [-1 , 1] .

Now the sequence $\sin(1) , \sin (2) , \sin (3) , \sin (4) ,....$ do not have a constant subsequence as the function $\sin(x)$ is periodic with period $2\pi$ so the terms of the sequence will repeat if $\sin (m) = \sin (n)$ that is m-n is a multiple of $2\pi$ which is not possible as m,n are integers.