let A = arctan(2) ------------------(1)tanA = 2 = 2 / 1
use these on a triangle with side opposite to A as 2, adjacent to A as 1
use pythagoras theorem to get the hypotenuse sqrt(5)
so sin A = 2 / sqrt(5)
using equation(1)
sin ( arctan(2) ) = 2 / sqrt(5) = [ 2*sqrt(5) ] / 5
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