What is cos(x+2pi)+cos(x-2pi)?

We'll transform the sum into a product using the
formula:

`cos a + cos b =
2cos[(a+b)/2]*cos[(a-b)/2]`

Let `a = x + 2pi and b = x -
2pi`

`a + b = x + 2pi + x - 2pi =
2x`

`a - b = x + 2pi - x +
2pi`

`a - b = 4pi`

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

`cos (x +
2pi) + cos (x - 2pi) = 2 cos x*cos 2pi`

`But cos 2pi =
1`

`cos (x + 2pi) + cos (x - 2pi) = 2 cos
x`

The sum can be calculated using the following
identities:

`cos (x + 2pi) = cos x*cos 2pi - sin x*sin
2pi`

Since `sin 2pi = 0` , the product `sinx*sin 2pi`  will
be 0.

`cos (x + 2pi) = cos
x`

`cos (x - 2pi) = cos x*cos 2pi + sin x*sin
2pi`

`cos (x - 2pi) = cos
x`

`cos (x + 2pi) + cos (x - 2pi) = cos x + cos x = 2 cos
x`

Therefore, the requested sum is `cos (x +
2pi) + cos (x - 2pi) = 2 cos x.`