Equate Derivatives to determine the 5th degree Taylor polynomial

Question

Equate Derivatives to determine the 5th degree Taylor polynomial for f(x)=sinx centered at a=2π/3.

Answer 1

f(x)=sin(x)≈f(2π/3)+f'((2π/3)(x-(2π/3))/1+f''((2π/3)(x-(2π/3))²/2+f'''((2π/3)(x-(2π/3)³)/6+f''''((2π/3)(x-(2π/3))⁴/24+f'''''((2π/3)(x-(2π/3))⁵/120

f((2π/3)=(√3)/2

f'((2π/3)=-1/2

f''((2π/3)=(-√3)/2

f'''((2π/3)=+1/2

f''''((2π/3)=f((2π/3)

f''''''((2π/3)=f'((2π/3)