Proving+a+product+sum+Identity+10-5

__**In Class**__
Today in class we talked about sums and products of two sinusoids and came to the realization that the product of 2Cos(X) and Cos(11X) is equal too the sum of Cos(12X) and Cos(10X)

__Prove It__
Cos(12X) and Cos(10X) can be re-written as the Cos(11X+X) and Cos (11X-X) which seems trivial and pointless but it furthers our proof by giving us a number which appears in both sides of the problem (i.e. 11x)

Cos(11X+X) + Cos (11X-X) can be re-written as Cos(11X)Cos(X)-Sin(11X)Sin(X) + Cos(11X)Cos(-X) -Sin(11X)Sin(-X) by the angle sum identity

This may seem useless now but because of that nice little negative X in the expression we can do a lot of canceling if we apply equations such as Cos(-X)=Cos(X) and Sin(-X)=-Sin(X) This allows us to express Cos(11X)Cos(X)-Sin(11X)Sin(X) + Cos(11X)Cos(-X) -Sin(11X)Sin(-X) as Cos(11X)Cos(X)-Sin(11X)Sin(X) + Cos(11X)Cos(X) + Sin(11X)Sin(X) This expression is very usefull because there are only two terms not four. The two terms are simplified as 2(Cos(11X)Cos(X)) +Sin(11X)Sin(X)-Sin(11X)Sin(X) Which simplifies to 2Cos(11X)Cos(x) which is what we started with.

__Generalize__
We can take what we just learned and apply it to give a general formula for writing Cos(AX)+Cos(BX) we can do what we did with 10,12 and 1 earlier and take an average between A and B and then add in the difference to keep the original value

math \cos(\frac{A+B}{2}x+\frac{A-B}{2}x) + \cos(\frac{B+A}{2}x+\frac{B-A}{2}x) math From there we can do what we did earlier using angle sum identities to show that the sum of the angles is

math 2\cos(\frac{A+B}{2}x)\cos(\frac{A-B}{2}) math