Answer
Not monotonic but bounded
Work Step by Step
$a_n=\:sin\left(\frac{n\times \pi }{6}\right)\:$
Let's calculate a few of the initial values to get an insight on the function as we cannot directly establish any relation between $a_n$ and $a_{n+1}$.
$a_1=\:sin\left(\frac{1\times \pi }{6}\right)\:=\:0.5\:$
$a_2=\:sin\left(\frac{2\times \:\pi \:}{6}\right)\:=sin\left(\frac{\pi \:}{3}\right)\:=\:0.866$
$a_3=\:sin\left(\frac{3\times \:\pi \:}{6}\right)\:=sin\left(\frac{\pi \:}{2}\right)\:=\:1$
$a_4=\:sin\left(\frac{4\pi \:}{6}\right)\:=sin\left(\frac{2\pi \:}{3}\right)\:=\:0.866$
Thus we see that the sequence first increases up till n=3 and then decreases. Hence it is not monotonic.
We know that with real numbers as arguments, sin(x) never outputs numbers outside the range of [-1, 1]. Hence the sequence is bounded from above and below.
