To find the MacLaurin series of #sin^2x# we can use the trigonometric identity:
#sin^2x = (1-cos2x)/2#
As the MacLaurin series of #cos t# is:
#cost = sum_(n=0)^oo (-1)^n t^(2n)/((2n)!)#
substituting #t = 2x# we have:
#cos2x = sum_(n=0)^oo (-1)^n (2x)^(2n)/((2n)!)#
and:
#sin^2x = 1/2 - 1/2sum_(n=0)^oo (-1)^n (2x)^(2n)/((2n)!)#
Extracting the term of index #n=0# from the sum we get:
#sin^2x = 1/2 - 1/2 - 1/2sum_(n=1)^oo (-1)^n (2x)^(2n)/((2n)!)#
and finally:
#sin^2x =sum_(n=1)^oo (-1)^(n+1) 2^(2n-1) x^(2n)/((2n)!) = x^2- 1/3x^4+...#
