Given
$F: \Bbb R^n$ $\mathbf x$ $\Bbb R$ $\rightarrow \Bbb R$, $n \in \Bbb N$,
$(x, t) = (x_1, x_2, ..., x_n, t) \rightarrow F(x, t)$,
$F(x, t) := \sin(\sum_{k=1}^n x_k - \sqrt nt)$,
$\Delta F$ $:=$ $\sum_{k=1}^n$ $d^2F \over d(x_k)^2$,
I would like to show that
$\Delta F$ $-$ $d^2F \over dt^2 $ = $0$.
Approach
I take a guess that this looks much more complicated than it actually is.
First, I calculate the necessary derivatives:
$dF\over d(x_k)^2$ $= \cos(\sum_{k=1}^n x_k - \sqrt n t)$
$d^2F \over d(x_k)^2$ $= - \sin(\sum_{k=1}^n x_k - \sqrt n t)$
$dF \over dt^2$ $= \cos(\sum_{k=1}^n x_k - \sqrt n t) * (-\sqrt n)$
$d^2F \over dt^2$ $=$ $n$ $*$ $\sin(\sum_{k=1}^n x_k - \sqrt nt) $
Recall that $sin(u)'=cos(u).u'$, $cos(u)'=-sin(u).u'$, $sin(-u)'=cos(u).(-u')$, and $\Delta F= \sum_{i=1}^n \partial_{x_i, x_i}F$.