How does it follow that $\frac{d^2\phi}{dx^2}=-\lambda\phi$ where our seperation constant $\lambda>0$ implies an exponential solution with imaginary exponents? Specifically $e^{\pm\imath\sqrt[]{\lambda}}$.
Then somehow if we want real independent solutions the choices $\cos(x\sqrt[]{\lambda})$ and $\sin(x\sqrt[]{\lambda})$? My text says $\cos(x\sqrt[]{\lambda})$ and $\sin(x\sqrt[]{\lambda})$ are linear combinations of $e^{\pm\imath\sqrt[]{\lambda}}$ but how? Does it have something to do with $e^{\imath\pi}=-1$?
It should be $e^{\pm ix\sqrt {\lambda}}$ not $e^{\pm i\sqrt {\lambda}}$.
This follows from the fact that $\sin a=\frac {e^{ia}-e^{-ia}} {2i}$ and $\cos a=\frac {e^{ia}+e^{-ia}} 2$.