Why write the solution of the harmonic oscillator form $$\psi=A\cos\omega_0 t+B\sin \omega_0t$$ is equal to writing form $$\psi=C_1e^{i\omega_0t}+C_2e^{-i\omega_0t}$$?
I would like to see how one implies the other or vice versa, the harmonic oscillator equation is: $$\frac{d^2}{dt^2}\psi+\omega_0^2\psi=0;\;\;\omega_0\in\mathbb{R}$$
$$\psi=A\cos\omega_0 t+B\sin \omega_0t$$
From $$\psi=C_1e^{i\omega_0t}+C_2e^{-i\omega_0t}$$ we can use Euler's equation: $$e^{ix}=\cos x+ i\sin x$$ to get: $$\psi=C_1*(\cos \omega_0t+ i\sin \omega_0t)+C_2*(\cos \omega_0t- i\sin \omega_0t)$$ $$\psi=(C_1+C_2) \cos \omega_0t+ (iC_1-iC_2)\sin \omega_0t$$ Finally define $$A= C_1+C_2 , B=iC_1-iC_2$$ Hence $$\psi=A\cos\omega_0 t+B\sin \omega_0t$$
Using Euler's formula on the second expression, we see
\begin{align*} \psi&=C_1e^{i\omega_0t}+C_2e^{-i\omega_0t} \\ &= C_1(\cos{\omega_0t}+i\sin{\omega_0t})+C_2(\cos(-\omega_0t)+i\sin(-\omega_0t)) \\ &= C_1(\cos{\omega_0t}+i\sin{\omega_0t})+C_2(\cos{\omega_0t}-i\sin{\omega_0t}) \\ &= (C_1+C_2)\cos{\omega_0t}+i(C_1-C_2)\sin{\omega_0t}\:. \end{align*}
If we then let $C_1=\frac{1}{2}(A-Bi),C_2=\frac{1}{2}(A+Bi)$, you get the first formula.
