Question

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

Answer 1

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...