# Maclaurin series help

1. Apr 26, 2005

### vabamyyr

i have trouble expandind f(x)= (sin2x)^2 into Maclaurin series
for sin(x) Maclaurin series is

$$\sum^{\infty}_{n=0} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$

probably the key is to change (sin2x)^2 into new shape. I found that

(sin2x)^2=2sin(2x^2), but that coefficent 2 is bothering me, what to do?

2. Apr 26, 2005

### Galileo

That relation does not hold (the left side is always positive, the right side isn't).

But you can use another identity:

$$\sin^2 x=\frac{1}{2}(1-\cos 2x)$$

3. Apr 26, 2005

### vabamyyr

$$\sin^2 2x=\frac{1}{2}(1-\cos 4x)=0.5-0.5cos4x$$

and now if i apply for cosx maclaurin series expansion considering the function f(x)= -0,5cos4x i get the right answer but im puzzled, where does the coefficent 0,5 go????? i dont have to count that??

Last edited: Apr 26, 2005
4. Apr 26, 2005

### mathman

The constant term of the series for sin2(2x) is 0.
The 0.5 cancels the -0.5 from the expansion of -0.5*cos(4x).

Last edited: Apr 26, 2005