use the taylor series format: #f(x)=f(a)/(0!)+(f'(a)(x-a))/(1!)+(f''(a)(x-a)^2)/(2!)+(f'''(a)(x-a)^3)/(3!)+...#
to find the taylor series with 3 nonzero terms, find #f(a), f'(a), f''(a), f'''(a)...#, where #a=0.5#, until you get 3 nonzero values for the nth derivative
#f(x)=sin(pix)rarrf(a)=sin(pi*0.5)=1# (used for first term)
#f'(x)=pi*cos(pix)rarrf'(a)=pi*cos(pi*0.5)=0#, so this term will equal zero and will not contribute to one of the 3 terms
#f''(x)=-pi^2*sin(pix)rarrf''(a)=-pi^2*sin(pi*0.5)=-pi^2# (used for second term)
#f'''(x)=-pi^3*cos(pix)rarrf'''(a)=-pi^3*cos(pi*0.5)=0# (again, this will not produce a new term)
#f''''(x)=pi^4*sin(pix)rarrf''''(a)=pi^4*sin(pi*0.5)=pi^4# (used for third term)
substitute the values in:
#f(x)~~1/(0!)+(0(x-0.5))/(1!)+(-pi^2(x-0.5)^2)/(2!)+(0(x-0.5)^3)/(3!)+(pi^4(x-0.5)^4)/(4!)#
#f(x)~~1+(-pi^2(x-0.5)^2)/(2!)+(pi^4(x-0.5)^4)/(4!)#
to approximate #sin(pi/2+pi/10)#:
#sin(pi/2+pi/10)=sin((6pi)/10)=sin((3pi)/5)#
if you let #x=3/5#, then #f(x)=sin((3pi)/5)#
substitue #x=3/5# into the taylor polynomial to get the approximation for #sin((3pi)/5)#:
#sin((3pi)/5)~~1+(-pi^2(3/5-0.5)^2)/(2!)+(pi^4(3/5-0.5)^4)/(4!)#
#sin((3pi)/5)~~0.951057849#
approximation for #sin(pi/2+pi/10): 0.951057849...#
actual value: #0.951056516...#
