How do you find the taylor series of #1 + sin(x)#?

The general form of the Taylor series is
#f(x) = Sigma_(n=0)^oo (f^(|n|) (a))/(n!)*(x-a)^n#
where #f^(|n|)# denotes the #n^(th)# derivative of #f#.

Set #f(x) = sin(x)#
and use #a=0#

Remember
#( dsin(x))/(dx) = cos(x)#
#(d cos(x))/(dx) = -sin(x)#
#sin(0) = 0#
#cos(0)=1#

Then the Taylor expansion of #1 + sin(x)#
becomes
#1 + x -(x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + (x^9)/(9!) ...#