sin(x) = x - x3/3! + x5/5! - x7/7! + ... + (-1)ix2i+1/(2i+1)! + ...
(where i=0,1,2,...). Let the nth sum, Sn(x), be given by
Sn(x) = x - x3/3! + x5/5! - x7/7! + ... + (-1)n-1x2n-1/(2n-1)!
and the error, or remainder, Rn(x), be given by
Rn(x) = (-1)nx2n+1/(2n+1)! + (-1)n+1x2n+3/(2n+3)! + ...
so that sin(x) = Sn(x) + Rn(x). We want to show that, if 2n >= |x|,
|Rn(x)| < |(-1)nx2n+1/(2n+1)!|
or, more simply,
|Rn(x)| < |x2n+1/(2n+1)!|.
First, assume that x >= 0.
Notice that, throughout the sine series, the ith term can be derived from the (i-1)th term by multiplying by (-1)x2/[(2i+1)(2i)]. For the terms in Rn(x), i >= n. Therefore, since 2n >= |x|, we have 2i >= |x|, and x2/[(2i+1)(2i)] < 1, which means the successive terms of Rn(x) are decreasing in absolute value. So Rn(x) can be expressed in the form
a1 - a2 + a3 - a4 + ... + (-1)i+1ai + ...
where each ai is positive, and
a1 > a2 > a3 > ... > ai > ...
Hence, we can apply the given result from calculus to conclude that
|Rn(x)| < |x2n+1/(2n+1)!|
and we are done.
If x < 0, this argument doesn't quite work, because Rn(x) will be of the form
-a1 + a2 - a3 + a4 - ...
when n is odd. In that case, let R'n(x) = -Rn(x), and use the same argument to show that
|R'n(x)| < |x2n+1/(2n+1)!|.
But |R'n(x)| = |Rn(x)|, so we can clearly conclude that for all values of x, the error is bounded as
|Rn(x)| < |nx2n+1/(2n+1)!|
which is what we want.
A similar argument works for cos(x).