f(x) = sinx , n=5 Find the n'th Maclaurin polynomial for the function.

Maclaurin series is a special case of Taylor series that is centered at a=0. The expansion of the function about 0 follows the formula:
f(x)=sum_(n=0)^oo (f^n(0))/(n!) x^n
 or
f(x)= f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4 +...
To determine the Maclaurin polynomial of degree n=5 for the given function f(x)=sin(x) , we may apply the formula for Maclaurin series.
To list f^n(x) , we may apply the derivative formula for trigonometric functions: d/(dx) sin(x) = cos(x)  and d/(dx) cos(x) = -sin(x) .
f(x) =sin(x)
f'(x) = d/(dx) sin(x)
           = cos(x)
f^2(x) = d/(dx) cos(x)
           = -sin(x)
f^3(x) = d/(dx) -sin(x)
           =-1*d/(dx) sin(x)
           = -1 * cos(x)
           = -cos(x)
f^4(x) = d/(dx) -cos(x)
            =-1*d/(dx) cos(x)
            = -1 * (-sin(x))
            = sin(x)
f^5(x) = d/(dx) sin(x)
          = cos(x)
Plug-in x=0 on each f^n(x) , we get:
f(0) =sin(0) =0
f'(0)= cos(0) =1
f^2(0)= -sin(0)=0
f^3(0)= -cos(0)=-1
f^4(0)= sin(0)=0
f^5(0)= cos(0)=1
Plug-in the values on the formula for Maclaurin series, we get:
sum_(n=0)^5 (f^n(0))/(n!) x^n= 0+1/(1!)x+0/(2!)x^2+(-1)/(3!)x^3+0/(4!)x^4+1/(5!)x^5
                       = 0+1/1x+0/2x^2-1/6x^3+0/24x^4+1/120x^5
                      = x-1/6x^3+1/120x^5
The Maclaurin polynomial of degree n=5 for the given function f(x)=sin(x) will be:
P_5(x)=x-1/6x^3+1/120x^5