Please explain the answer

14. The Fourier sine series of f(x) - 1, 0 A 1 on [0, 2] is (a) - cos (b) 2 sin nn ("2)1 2 [1 - (-1)" sin n=1 (c) 1 - (-1) sin(nnx) (d) 1 - cos(2nm ) sin(nar) (e) None of these n=1 n=]