We know that the maclaurin series for #sin(x) = sum_(n=0)^∞ ((-1)^n * (x)^(2n+1))/((2n+1)!)#
Thus the series #sin(3x)# is identical except we replace all #x#'s with#3x#, the series then becomes #sin(3x) = sum_(n=0)^∞ ((-1)^n * (3x)^(2n+1))/((2n+1)!)#
We know the first term for the maclaurin series for #sin(x)# is #x#, so the first term for #sin(3x)# is #3x#, we can check be inputting #n=0# into the series.
Next, we can pull out the first term out of the summation, we must shift the index by 1 because we removed one term. The series now becomes #sin(3x) = 3x + sum_(n=1)^∞ ((-1)^n * (3x)^(2n+1))/((2n+1)!)#
Now we can easily compute the series for #sin(3x) - x#, all we need to do is subtract #x# from the above series:
#sin(3x) - x = 3x + (sum_(n=1)^∞ ((-1)^n * (3x)^(2n+1))/((2n+1)!)) - x#
This becomes #sin(3x) - x = 3x - x + (sum_(n=1)^∞ ((-1)^n * (3x)^(2n+1))/((2n+1)!))#
So we finally get, #sin(3x) - x = 2x + (sum_(n=1)^∞ ((-1)^n * (3x)^(2n+1))/((2n+1)!))#
