Find the Taylor formula up to n=3 for f(x)=sin(x) at X0=pi/3:
sin(x)=a0+a1(x-pi/3)+a2(x-pi/3)^2+a3(x-pi/3)^3+R3
Use this formula to find the approximate value of sin(58 degees)
Just plug and chug. You know that
f = sin
f' = cos
f" = -sin
f(3) = -cos
So,
f(x) = f(π/3) + f'(π/3)(x-π/3) + f"(π/3)/2! (x-π/3)^2 + f(3)(π/3)/3! (x-π/3)^3 + ...
f(x) = √3/2 + (1/2)(x-π/3) - (√3/2)/2! (x-π/3)^2 - (1/2)/3! (x-π/3)^3 + ...
f(x) = √3/2 + (1/2)(x-π/3) - (√3/4)(x-π/3)^2 - (1/12)(x-π/3)^3 + ...