Show that for any irrational @ the limit

Show that for any irrational \alpha the limit
\displaystyle\mathop {\lim }\limits_{n \to \infty } \sin \left( {n\alpha \pi } \right)
does not exist.
Solution. If the limit existed then we would get
\displaystyle 0 = \mathop {\lim }\limits_{n \to \infty } \left( {\sin \left( {\left( {n + 2} \right)\alpha \pi } \right) - \sin \left( {n\alpha \pi } \right)} \right) = 2\sin \left( {\alpha \pi } \right)\mathop {\lim }\limits_{n \to \infty } \cos \left( {\left( {n + 1} \right)\alpha \pi } \right)
and consequently, \mathop {\lim }\limits_{n \to \infty } \cos \left( {n\alpha \pi } \right) = 0. Similarly,
\displaystyle0 = \mathop {\lim }\limits_{n \to \infty } \left( {\cos \left( {\left( {n + 2} \right)\alpha \pi } \right) - \cos \left( {n\alpha \pi } \right)} \right) = - 2\sin \left( {\alpha \pi } \right)\mathop {\lim }\limits_{n \to \infty } \cos \left( {\left( {n + 1} \right)\alpha \pi } \right)
which is impossible because \sin^2x+\cos^2x=1 for all x \in \mathbb R. Therefore the limit does not exist.
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