## Some trigonometric, exponential thing?

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

Try Euler's formula
what do you get?
 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?

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## Some trigonometric, exponential thing?

 Quote by M. next it would be: A(coskx +isinkx)+B(coskx-isinkx) which's (A+B)coskx+i(A-B)sinkx
so B' = i(A-B) …
i told you they won't both be real!
 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.

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