First find the solution to the linear homogeneous wave equation with wave speed $1$ and with initial conditions $u(x, 0) = \sin x$, $u_t(x, 0) = 0.$ Then calculate $u_t(0, t).$
I'm stuck on this question by calculating $u_t(0, t).$ So first the general solution for $u_{tt} = u_{xx}$ is $$u(x, t) = \frac{1}{2}[\phi(x + t) + \phi(x - t)] + \frac{1}{2} \int_{x - t}^{x + t} \psi(s)\,{\rm d}s$$ which is d’Alembert’s formula. With initial condition $u(x, 0) = \sin x$, $u_t(x, 0) = 0.$ It gives
$$u(x, t) = \frac{1}{2}[\phi(x) + \phi(x)] + \frac{1}{2} \int_{x}^{x}0\,{\rm d}s,$$so
$$u(x, t) = \frac{1}{2}[\sin(x + t) + \sin(x - t)].$$
This is where I got stuck, because I'm not sure what to do with $u_t(0,t)$. Any suggestions?
If $u_t(x,0) = 0$, what is the issue? We don't care about $u_t(0,t)$ when applying d'Alembert's formula anyway. You can even simplify what you want further to end up with $$u(x,t) = \sin x \cos t.$$And indeed, $u(x,0) = \sin x$ because $\cos 0 = 1$, and $u_t(x,0) = 0$ because $\sin 0 = 0$. Moreover $$u_{tt}(x,t) - u_{xx}(x,t) = -\sin x\cos t - (-\sin x \cos t) = 0.$$So $u_t(0,t) = -\sin 0 \sin t = 0$. You were essentially done without realizing.