Find the Maclaurin series for f (x) = sin^2 (x)

use sin^2 (x) = 1/2 * (1-cos2x)

cos 2x - use def cosx from table

f (x) = sin^2 (0) f' (x) = 2*sin (x) * cos (x)

therefore f' (x) = sin (2x)

f'' (x) = 2cos (2x)

f''' (x) = - 4sin (2x) = - 4*f' (x)

putting it into maclaurin's series

We get,

0+0 + 2x^2/2! + 0 ...