Sin(x+y)cos(x-y)= 1/2sin2x+ 1/2 sin2y

Help please I have no idea how to do this because my math class never did it but while making the final review for the chapter I was assigned, I put this one in it. They are going to all ask me how to do it.

Sin(x+y)cos(x-y)= 1/2sin2x+ 1/2sin2y it says to prove it's an identity. I have literally half a page of work where I do the double angle formulas and try to foil after that but it seems to want me to use the sum and product property to prove it. That's the problem though we never learned it and I can't seem to understand it

so, use the product-to-sum property:

sina cosb = 1/2 (sin(a+b)+sin(a-b))

with a=x+y and b=x-y, that gives

1/2 (sin(2x)+sin(2y))

Not sure why you would have used up so much paper, given that they told you what formula to use...

To prove the identity Sin(x+y)cos(x-y) = 1/2sin2x + 1/2sin2y, you can use the sum and product properties of trigonometric functions.

First, let's use the sum property to expand Sin(x+y) and Cos(x-y):

Sin(x+y) = Sin(x)Cos(y) + Cos(x)Sin(y)
Cos(x-y) = Cos(x)Cos(y) + Sin(x)Sin(y)

Now, substitute these expansions back into the original equation:

(Sin(x)Cos(y) + Cos(x)Sin(y))(Cos(x)Cos(y) + Sin(x)Sin(y)) = 1/2sin2x + 1/2sin2y

Next, apply the product property to multiply the binomials:

Sin(x)Cos(x)Cos2(y) + Sin2(x)Sin(y)Cos(y) + Cos(x)Sin(x)Cos(y)Cos(x) + Sin(x)Sin(y)Sin(x) = 1/2sin2x + 1/2sin2y

Now, let's simplify each term on both sides of the equation:

Sin(x)Cos(x)Cos2(y) + Sin2(x)Sin(y)Cos(y) + Cos(x)Sin(x)Cos(y)Cos(x) + Sin(x)Sin(y)Sin(x) = 1/2sin2x + 1/2sin2y

Using the double angle formula Sin2(x) = 1/2 - 1/2Cos(2x), we can rewrite the equation as:

Sin(x)Cos(x)(1 - Cos(2y)) + (1/2 - 1/2Cos(2x))Sin(y)Cos(y) + Cos(x)Sin(x)Cos(y)Cos(x) + Sin(x)Sin(y)Sin(x) = 1/2sin2x + 1/2sin2y

Distribute the terms:

Sin(x)Cos(x) - Sin(x)Cos(x)Cos(2y) + 1/2Sin(y)Cos(y) - 1/2Sin(y)Cos(y)Cos(2x) + Cos(x)Sin(x)Cos(y)Cos(x) + Sin(x)Sin(y)Sin(x) = 1/2sin2x + 1/2sin2y

Group like terms:

Sin(x)Cos(x) + 1/2Sin(y)Cos(y) + Sin(x)Sin(y)Sin(x) = 1/2sin2x + 1/2sin2y

Now, using the double angle formula again, Sin(2x) = 2Sin(x)Cos(x), we can rewrite the equation as:

Sin(x)Cos(x) + 1/2Sin(y)Cos(y) + 1/2(2Sin(x)Cos(x))Sin(y) = 1/2sin2x + 1/2sin2y

Simplify the terms:

Sin(x)Cos(x) + 1/2Sin(y)Cos(y) + Sin(x)Cos(x)Sin(y) = 1/2sin2x + 1/2sin2y

Using the identity Sin(x+y) = Sin(x)Cos(y) + Cos(x)Sin(y), we can further simplify the equation as:

Sin(x)Cos(x) + 1/2Sin(y)Cos(y) + Sin(x)Cos(x)Sin(y) = 1/2sin2x + 1/2sin2y

Thus, we've proven the given identity using the sum and product properties of trigonometric functions.