Using Taylor theorem prove that x-((x^3)/6)0

Question

Using Taylor theorem prove that x-((x^3)/6)<sinx<x-((x^3)/6)+((x^5)/120) for x>0

Patrick B. · Accepted Answer

When expanded in a Taylor series, the sine function is

f(x) = sin(x) = x - x^3/6 + x^5/5! - x^77! + x^9/9! - x^11/11! + .... + (-1)^(n)*x^(2n+1)/(2n+1)! as n->infinity

where each term is LESS than the previous term in absolute value;

Indeed, x - x^3/6 < f(x) because of the sum of the terms beginning with n=3

and f(x) < x - x^3/6 + x^5/5! because of the terms beginning with n=4

Using Taylor theorem prove that x-((x^3)/6)<sinx<x-((x^3)/6)+((x^5)/120) for x>0