The Maclaurin series is given by:
#f(x) = sum_(n=0)^oof^n(0)/(n!)x^n#
where #f^n(0)# denotes the #n#th derivative of #f(x)# evaluated at 0.
So as you have stated the list of the derivatives are as follows:
#f(x) = sin(x +pi/5) #
#f'(x) = cos(x +pi/5) #
#f''(x) = -sin(x +pi/5) #
#f'''(x) = -cos(x +pi/5) #
Evaluating these at #x=0# gives us the following list (it turns out these have the following exact values:
#f(0)=sin(pi/5)=1/4sqrt(10-2sqrt5)#
# f'(0) = cos(pi/5)=1/4sqrt(1+sqrt(5))#
#-> f''(0) = -sin(pi/5)=-1/4sqrt(10-2sqrt5)#
#-> f'''(0)=-cos(pi/5)=-1/4sqrt(1+sqrt(5))#
This will allow us to find the first 4 terms of the Maclaurin series (if you wish to find more terms and therefore have a more accurate series simply find further derivatives).
So from the definition of the Maclaurin series above:
#f(x) = sum_(n=0)^oof^n(0)/(n!)x^n#
#=(f(0))/(0!)x^0+(f'(0))/(1!)x^1+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+...#
#=f(0)+f'(0)x+(f''(0))/2x^2+(f'''(0))/(6)x^3+...#
It is now a simple matter of substituting the values worked out from the list above in place of #f^n(x)# like so:
#f(x)=sin(pi/5)+cos(pi/5)x-sin(pi/5)x^2/2-cos(pi/5)x^3/6+...#
You could stop here or continue and replace these trig terms with their exact values like so:
#=1/4sqrt(10-2sqrt5)+1/4sqrt(1+sqrt(5))*x-1/4sqrt(10-2sqrt5)*x^2/2-1/4sqrt(1+sqrt(5))*x^3/6+...#
#=1/4sqrt(10-2sqrt5)+1/4sqrt(1+sqrt(5))*x-1/8sqrt(10-2sqrt5)*x^2-1/24sqrt(1+sqrt(5))*x^3+...#
Of course, as stated above, a Maclaurin series generally carries on to infinity. If you wish to find further terms you must calculate further derivatives of #f(x)# until you you have reached your goal.
