Answer to Question #283841 in Calculus for Tushal

Question

Using Taylor's theorem prove that. x-(x^3/6)<sinx<x-(x^3/6)+(x^5/120)

Answer 1

The infinite Maclaurin series for sine is well known (and it has alternating signs from term to term). (Special case of Taylor series for a=0)

 

Referring to your 3-part inequality, the left side is that series truncated to 2 terms; the right side is that series truncated to 3 terms.

 

Since, in the case of the left side, the omitted terms decrease in absolute value, uniformly (abs. values are successively smaller & smaller), and the first of those omitted terms would be positive, the left-side 'estimate' is necessarily less than the the true sin x value.

 

Since, in the case of the right side, the omitted terms decrease in absolute value, uniformly, and the first of those omitted terms would be negative, the right-side 'estimate' is necessarily greater than the the true sin x value.