- #1
M. next
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How can we say:
f(x)=A'sin(kx)+B'cos(kx)
or equivalently
f(x)=Ae[itex]^{ikx}[/itex]+Be[itex]^{-ikx}[/itex]??
How are these two equivalent knowing that e[itex]^{ix}[/itex]=cosx+isinx
I don't get this?
f(x)=A'sin(kx)+B'cos(kx)
or equivalently
f(x)=Ae[itex]^{ikx}[/itex]+Be[itex]^{-ikx}[/itex]??
How are these two equivalent knowing that e[itex]^{ix}[/itex]=cosx+isinx
I don't get this?