\lim_{n\rightarrow\infty}\frac{|a_{n+1}|}{|a_n|}=\lim_{n\rightarrow\infty}\frac {2^{n+1}(sin(\frac 1{2^{n+1}}))^2}{{2^n(sin(\frac 1{2^{n}}))^2}}=\lim_{n\rightarrow\infty}\frac {2(\frac 1{2^{n+1}})^2}{(\frac 1{2^{n}})^2}= \frac 12