Question: This should be simple but I am horrible at math.

Anyways I forgot basic math properties and when I try to work it out in more simple terms I can't make sense of anything. Anyways I have to factor $(\sin x+1)^2 - (\sin x-1)^2$

So I figure that I just do $(\sin x+1)(\sin x+1)$ for the first term and then $(\sin x-1)(\sin x-1)$ for the second I wasn't too sure on the second one but I figured that since the negative sign was outside the expression that I just leave it there, do my math and then apply it. That makes sense to me with what I know of how parentheses work.

Anyways I got the wrong answer and this is where I think it is incorrect, my book does some very strange stuff that I do not understand like adding instead of multiplying the terms.

Answer: 
Any time you have a **difference of squares**,
$$a^2 - b^2$$
it can be factored as
$$a^2-b^2 = (a+b)(a-b).$$

Here you have a difference of squares, with $a=\sin x+1$ and $b=\sin x -1$, so you can factor it this way, and then simplify to see what you get. Note that this factorization is completely independent of the fact that you are dealing with trigonometric expressions; it's just simple algebra.
Arturo's answer is the 'slickest' way to get to the answer, but if you wanted to be more pedestrian you could just multiply out the brackets one at a time, and then do the necessary simplifications:

$$\begin{align}
(\sin x + 1)^2 - (\sin x - 1)^2
&= (\sin^2 x + 2\sin x + 1) - (\sin^2 x - 2\sin x + 1) & \textrm{(multiply out the brackets)} \\
&= \sin^2 x + 2\sin x + 1 - \sin^2 x + 2\sin x - 1 & \textrm{(take the negative of the second bracket)} \\
& = 2\sin x + 2\sin x & \textrm{(the }\sin^2x\textrm{ and 1 terms cancel)} \\
& = 4\sin x & \textrm{(collect everything together)}
\end{align}
$$