Which is not right ... I'm pretty sure I'm just making a stupid mistake with my product rule, but I've tried it a couple times and I keep doing the same thing.

Does anyone see why I'm getting 2's in front of everything when I shouldn't be?

You are differentiating with respect to x; that means that cos(a) and sin(a) are just constants. Differentiate again, treat them as constants, and you'll get the correct sum formula for cos(x + a). Double-angle formula would be for cos2x.

hmm ... I don't know if you're right.
I've seen the double angle formula with two separate variables.
Plus it shouldn't matter whether it's a constant or a variable, the formula should still work out if the end result is supposed to be in terms of both x and a.

Re-word your question. Prove that for any constant a,

cos(x+a)=cos(x)cos(a)-sin(x)sin(a)

Do you see how this is equivalent?

Basically, the error that you made is that even if a is a variable, not a constant, it's independent of x as an input. So when you differentiate with respect to x, you treat the a as constant. This is something covered in Calculus III classes usually (multivariable calc)

Yes, you do use two different values for the sum formulas. However, you need to differentiate to get the sum formula for cosine, and you are using d/dx which is differentiating with respect to x. Did you notice that when you differentiated the left side, d/dx sin(x + 1), you treated a as a constant? Although a is a constant, you can let it be any value when using the formula, just like when you choose any value for x.

My friends, first of all "x" & "a" can be both variable or can be a variable "x" and a constant "a" in the expression: Sin(x+a) = SinxCosa + CosxSina. To prove, try solving:
Sin(30+5) = Sin30Cos5 + Cos30Sin5 and youll find out the formula works. Now, starting from the fact that they can both variable or a constant and a variable, we could think of two possible solutions to this problem which of course will lead to the same result: Cos(x+a) = CosxCosa - SinxSina.

Lets try the first assumption, i.e., x is a variable and a as a constant:

Sin(x+a) = SinxCosa + CosxSina--->differentiate both sides with respect to x.
you'll get: cos(x+a) = [sinx*(0) + cosa*cosx] + [cosx*(0) + sina*(-sinx)]-->notice the derivative of cosa = 0 because the derivative of a constant is 0, likewise the derivative of sina.

Simplifying, we get: Cos(x+a) = CosxCosa - SinxSina.--->which solves the problem.