2016/098) Let $a = 5^{1000}\sin(1000\alpha)$ where $\sin(\alpha) = \frac{3}{4}$ prove that $a\in Z$

because $\sin \alpha = \frac{3}{5}$ we have $\cos \alpha = \frac{\pm 4}{5}$
we have $\sin(1000\alpha) = Im(\cos(1000\alpha) + i\sin (1000\alpha)) = Im(e^{1000\alpha i})$
$=Im((e^{\alpha i})^{1000})  = Im((\cos \alpha + i\sin\alpha)^{1000}) = Im ( \frac{\pm 4}{5} + i \frac{3}{5})^{1000}$
hence $a = Im (\pm 4 + 3i)^{1000} = Im(\sum_{n=0}^{1000}{1000 \choose b} (\pm 4)^n * (3i)^{1000-n}$
A is imaginary part of the above and as each element is integer we have a is integer