How do you derive e^ikx = cos(kx) + i·sin(kx) from Euler's formula?

[SpartanG345]

Eulers formula says that e^ix = cosx + isinx

but in my textbook there's another formula its e^ikx = coskx + isinkx
i still can't figure out how they got that. Is this still eulers formula? and how do u get it in that form

 

[Feldoh]

e^iy = cosy + isiny

Let y = kx

e^ikx = coskx + isinkx