Trigonometric function

f(x)=3sin(2pi*x)

Evaluate A=f(1/2011)-f(2012/2011)

Observe that 2012/2011 = 1 + (1/2011) so

f(2012/2011) = 3 sin(2π(1 + (1/2011)) = 3 sin(2π + 2π/2011)

Using the identity for the sine of a sum,

sin(2π + 2π/2011) = sin(2π) cos( 2π/2011) + cos(2π) sin(2π/2011)
  = 0 * cos( 2π/2011) + 1 * sin(2π/2011)
  = sin(2π/2011)

Therefore, A = 0.