Ans : Note that:
sin(x) = sum(n=0 to infinity) [(-1)^n * x^(2n + 1)]/(2n + 1)!.
Then, since the series is alternating, the error in the approximation found by taking the first n terms of the series is no bigger than the n+1'th term. In other words:
E ≤ a_n+1 = x^(2n + 3)/(2n + 3)!.
(Note that a_n does not include (-1)^n, the alternating part)
We need that 1/6 ≤ x^(2n + 3)/(2n + 3)!. Given |x| < 1, n = 2 will be the least integer solution. Thus, we need 2 + 1 = 3 terms.
