Taylor series expansion of $\sin^2(x)$

Question

I'm looking at a solution and I'm having a hard time understanding this equality (I'm not great at Sigma notation),

$ \sin^2(x) = \dfrac{1-\cos(2x)}{2}=\dfrac{1}{2}\left[1-\displaystyle\sum_{n=0}^\infty\dfrac{(-1)^n(2x)^{2n}}{(2n)!}\right] = \dfrac{1}{2}\left[1-1-\displaystyle\sum_{n=1}^\infty\dfrac{(-1)^n(2x)^{2n}}{(2n)!}\right] $

It's unclear to me how to go from this

$ \dfrac{1}{2}\left[1-\displaystyle\sum_{n=0}^\infty\dfrac{(-1)^n(2x)^{2n}}{(2n)!}\right] $

to this

$ \dfrac{1}{2}\left[1-1-\displaystyle\sum_{n=1}^\infty\dfrac{(-1)^n(2x)^{2n}}{(2n)!}\right] $

Answer 1

We know that $$\cos (y) =\sum_{n=0}^\infty \frac{(-1)^n y^{2n}}{(2n)!}= 1-\frac{y^2}{2!}+\frac{y^4}{4!}+\ldots$$

from Taylor series of cosine.

We just replace $y=2x$.

Answer 2

Another possible way is to differentiate $\sin^2x$ and observe that$$[\sin^2x]'=2\sin x\cos x=\sin 2x$$

Thus, using the taylor series for $\sin x$ gives$$\sin 2x=\sum\limits_{n\geq0}(-1)^n\frac {(2x)^{2n+1}}{(2n+1)!}=2\sum\limits_{n\geq0}(-1)^n\frac {4^nx^{2n+1}}{(2n+1)!}$$

Now integrate with respect to $x$ to get the expansion!

Answer 3

The sum initially goes from $n=0$ to $n=\infty$. In the second line, the first term ($n=0$), which is $1$, is expanded apart, and now the sum starts at $n=1$. And don't be confused by the minus "-" in front :)