The general expression of the Taylor series of #f(x)# around #x=x_0# is:
#f(x) = sum_(n=0)^oo f^((n))(x_0)/(n!)(x-x_0)^n#
For #f(x) = sinx# and #x_0 = pi/2# we have accordingly:
#f^((0))(pi/2) = sin(pi/2) = 1#
#f^((1))(pi/2) = cos(pi/2) = 0#
#f^((2))(pi/2) = -sin(pi/2) = 1#
#f^((3))(pi/2) = -cos(pi/2) = 0#
Then:
#sinx = 1-1/2(x-pi/2)^2+o((x-pi/2)^4)#
#sinx = 1-pi^2/8 +(pix)/2-x^2/2+o((x-pi/2)^4)#
We can also note that:
#sinx = sinxsin(pi/2)+cosxcos(pi/2) = cos(x-pi/2)#
so we can use the MacLaurin expansion of #cosx# substituting #(x-pi/2)# for #x# and have that around #x=pi/2#:
#sinx = sum_(n=0)^oo (-1)^n/((2n)!) (x-pi/2)^(2n)#
