# How transform cos x-cos y+sin(x+y) in product?

### 1 Answer | Add Yours

You need to convert first the difference `cos x - cos y` in a product, using the following formula, such that:

`cos x - cos y = 2 sin((x+y)/2)sin((y-x)/2)`

Replacing `2 sin((x+y)/2)sin((y-x)/2)` for `cos x - cos y` in expression, yields:

`2 sin((x+y)/2)sin((y-x)/2) + sin(x+y)`

You need to use the double angle identity, such that:

`sin 2alpha = 2 sin alpha*cos alpha`

Reasoning by analogy, yields:

`sin(x+y) = 2 sin((x+y)/2)cos((x+y)/2)`

Replacing `2 sin((x+y)/2)cos((x+y)/2)` for `sin(x+y)` in expression, yields:

`cos x - cos y + sin(x+y) = 2 sin((x+y)/2)sin((y-x)/2) + 2 sin((x+y)/2)cos((x+y)/2)`

Factoring out `2 sin((x+y)/2)` yields:

`cos x - cos y + sin(x+y) = 2 sin((x+y)/2)(sin((y-x)/2) + cos((x+y)/2))`

You need to use the following identity, such that:

`sin alpha = cos(pi/2 - alpha)`

Reasoning by analogy, yields:

`sin((y-x)/2) = cos(pi/2 - y/2 + x/2)`

Replacing `cos((pi + x - y)/2)` for `sin((y-x)/2)` yields:

`cos x - cos y + sin(x+y) = 2 sin((x+y)/2)(cos((pi + x - y)/2) + cos((x+y)/2))`

Converting the sum `cos((pi + x - y)/2) + cos((x+y)/2)` into a product, yields:

`cos((pi + x - y)/2) + cos((x+y)/2) = 2cos((pi + x - y + x + y)/4)cos((pi + x - y - x - y)/4)`

`cos((pi + x - y)/2) + cos((x+y)/2) = 2cos((pi + 2x)/4)cos((pi - 2y)/4)`

Replacing `2cos((pi + 2x)/4)cos((pi - 2y)/4)` for` cos((pi + x - y)/2) + cos((x+y)/2)` yields:

`cos x - cos y + sin(x+y) = 4 sin((x + y)/2)cos((pi + 2x)/4)cos((pi - 2y)/4)`

**Hence, converting the given expression into a product, using trigonometric identities, yields **`cos x - cos y + sin(x+y) = 4 sin((x + y)/2)cos((pi + 2x)/4)cos((pi - 2y)/4).`