Some trigonometric, exponential thing?

1. Nov 3, 2012

M. next

How can we say:

f(x)=A'sin(kx)+B'cos(kx)

or equivalently

f(x)=Ae$^{ikx}$+Be$^{-ikx}$??

How are these two equivalent knowing that e$^{ix}$=cosx+isinx

I don't get this?

2. Nov 3, 2012

tiny-tim

Hi M. next!
They won't both be real.

Try Euler's formula

what do you get?

3. Nov 3, 2012

M. next

it would be: A(coskx +isinkx)+B(coskx-isinkx)
which's (A+B)coskx+i(A-B)sinkx
.. A'coskx+iB'sinkx
where's did the "i" go?

4. Nov 3, 2012

tiny-tim

so B' = i(A-B) …

i told you they won't both be real!

5. Nov 18, 2012

M. next

Sorry, i didn't check the site from since, I had some connection difficulties.
So, my final question, can this be done? Is the exponential form an alternative for the known trigonometric one?
And why do I use it? Why not keep it in trigonometric form. I am working on potential wells, free particles and so, if this information would help you answer my question.

6. Nov 18, 2012

tiny-tim

Hi M. next!
Yes, they're equally valid alternatives.

You use cos and sin, or real exponentials, if you're only interested in real solutions,

but you use complex exponentials if you're interested in complex solutions.

7. Nov 18, 2012

M. next

Thanks, am grateful