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 with3x, 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)!))
