# Maclaurin series for sin^2(x)

1. May 23, 2013

### John112

since the maclaurin series for sin x is alternating in sign (EQ1) so when you square it to get sin$^{2}$(x) (EQ2) the (-1)$^{n}$ should become (-1)$^{2n}$ (EQ3) which can be simplified down to (EQ4), but when i checked that series at wolframalpha the series was still alternating like: Why is that? So when we square it do we ignore squaring the (-1)$^{n}$ and put that after we're done squaring the series?

In the above post when I say EQ#, I'm referring to these equations

Last edited by a moderator: Apr 18, 2017
2. May 23, 2013

### micromass

Staff Emeritus
You seem to think that $\left(\sum \alpha_n\right)^2 = \sum \alpha_n^2$. This is of course false. It's the same thing as saying that $(x + y)^2 = x^2 + y^2$.

3. May 23, 2013

### John112

I can now see why sin$^{2}$(x) should be alternating, but how would I then simplify this algerbracially? how would I simplify $\left(\sum \alpha_n\right)^2$ ? Is using the half angle property for sin$^{2}$(x) my only method?

Last edited: May 24, 2013
4. May 24, 2013

### Staff: Mentor

It's probably the easiest way to go.

5. May 24, 2013

### micromass

Staff Emeritus
The half-angle property is indeed the easiest way. But there also is a explicit way to multiply two series. This is called the Cauchy product: http://en.wikipedia.org/wiki/Cauchy_product So you can solve it with this too, but this is a lot more complicated.

Another way is to explicitely find the derivatives of $\sin^2(x)$ and see if you can find a pattern. But this is also not a very simple way to go.