Answer
By differentiation we show that $u\left( {t,x} \right) = \sin \left( {\alpha ct + \beta } \right)\sin \left( {\alpha x} \right)$ satisfies the wave equation:
$\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {c^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}$
Work Step by Step
We have $u\left( {t,x} \right) = \sin \left( {\alpha ct + \beta } \right)\sin \left( {\alpha x} \right)$.
Step 1. Differentiation with respect to $t$
$\frac{{\partial u}}{{\partial t}} = \alpha c\cos \left( {\alpha ct + \beta } \right)\sin \left( {\alpha x} \right)$
$\frac{{{\partial ^2}u}}{{\partial {t^2}}} = - {\alpha ^2}{c^2}\sin \left( {\alpha ct + \beta } \right)\sin \left( {\alpha x} \right)$
$ = {c^2}\left( { - {\alpha ^2}\sin \left( {\alpha ct + \beta } \right)\sin \left( {\alpha x} \right)} \right)$
Step 2. Differentiation with respect to $x$
$\frac{{\partial u}}{{\partial x}} = \alpha \sin \left( {\alpha ct + \beta } \right)\cos \left( {\alpha x} \right)$
$\frac{{{\partial ^2}u}}{{\partial {x^2}}} = - {\alpha ^2}\sin \left( {\alpha ct + \beta } \right)\sin \left( {\alpha x} \right)$
Since $\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {c^2}\left( { - {\alpha ^2}\sin \left( {\alpha ct + \beta } \right)\sin \left( {\alpha x} \right)} \right)$, we conclude that $u\left( {t,x} \right) = \sin \left( {\alpha ct + \beta } \right)\sin \left( {\alpha x} \right)$ satisfies the wave equation:
$\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {c^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}$
