By replacing a with 2x in the Maclaurin series for sin x, we can see that...

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By replacing a with 2x in the Maclaurin series for sin x, we can see that sin(2x) == n=0 (-1) 22n+1 2+1 (2n+1)! In this question, you will find another path towards determining the Maclaurin series for sin(2x). (a) Use the fact that sin x 1 - cos(2x) to obtain the Maclaurin series for sin x. Note that 2 COS x = (-1)x2n (2n)! (b) Use the fact that d dx sin2 x 2 sin x cos x = = sin(2x) to obtain the Maclaurin series for sin(2x). By replacing a with 2x in the Maclaurin series for sin x, we can see that sin(2x) == n=0 (-1) 22n+1 2+1 (2n+1)! In this question, you will find another path towards determining the Maclaurin series for sin(2x). (a) Use the fact that sin x 1 - cos(2x) to obtain the Maclaurin series for sin x. Note that 2 COS x = (-1)x2n (2n)! (b) Use the fact that d dx sin2 x 2 sin x cos x = = sin(2x) to obtain the Maclaurin series for sin(2x).