Some trigonometric, exponential thing?

[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?

 

[tiny-tim]

Hi 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? ​

 

[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?

 

[tiny-tim]

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! ​

 

[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.

 

[tiny-tim]

Hi M. next!

Is the exponential form an alternative for the known trigonometric one?
And why do I use it? Why not keep it in trigonometric form.

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. ​

 

[M. next]

Thanks, am grateful