Expressing solution of this wave equation problem in different Fourier expansion

Question

I have managed to solve the wave equation $u_{tt} = c^{2} u_{xx}$ on the interval $[0,L]$ for $t > 0$, and subject to initial conditions $u(x,0) = f(x)$ and boundary conditions $u_{t}(x,0) = g(x)$. My solution is:

$$u(x,t) = \sum_{n=1}^{\infty} \left(\sin{\frac{n\pi{x}}{L}} \left( A_{n} \cos{\frac{n\pi{ct}}{L}} + B_{n}\sin{\frac{n\pi{ct}}{L}}\right) \right)$$

and I determined the Fourier coefficients to be $$\begin{cases} A_{n} = \displaystyle\frac{2}{L} \displaystyle\int_{0}^{L} f(x) \sin{\frac{n\pi{x}}{L}} \\ B_{n} = \displaystyle\frac{n\pi{c}}{L} \displaystyle\int_{0}^{L} g(x) \sin{\frac{n\pi{x}}{L}}\end{cases} $$

What I can't show is that for $u_{n}(x,t)$, the $n$-th term of our solution, we can write

$$u_{n}(x,t) = C_{n} \sin{\left( \frac{n\pi{ct}}{L} + \psi_{n} \right)} \sin{\frac{n\pi{x}}{L}}$$

I'm pretty sure that $C_{n} = \sqrt{A_{n}^{2} + B_{n}^{2}}$, since the Fourier coefficient $C_{n}$ should denote the amplitude.

Answer 1

$$ A\cos \omega t+B \sin \omega t=\\ \sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}}\cos \omega t+\frac{B}{\sqrt{A^2+B^2}}\sin \omega t\right)=\\ \sqrt{A^2+B^2}\left(\sin \psi\cos\omega t+\cos\psi\sin\omega t\right)=\\ \sqrt{A^2+B^2}\sin(\omega t +\psi) $$

Answer 2

Suppose $a$, $b$ are real numbers. Then $a^{2}+b^{2}=1$ iff $a=\sin(\psi)$, $b=\cos(\psi)$ some $\psi\in[0,2\pi)$. In this case, $\psi\in [0,2\pi)$ is unique.