By definition we have
$\sin(z) = \frac{1}{2\imath} \cdot \left(e^{\imath \, z}-e^{-\imath \, z}\right) $
Since the sum of two analytic functions is analytic, it suffices to show that $z \mapsto e^{\imath \, z}$ and $z \mapsto e^{-\imath \, z}$ are analytic. Let $z:=x+ \imath \, y$ ($x,y \in \mathbb{R}$), then
$e^{\imath \, z} = e^{\imath \, x} \cdot e^{-y} = \underbrace{e^{-y} \cdot \cos(x)}_{=:u(x,y)}+\imath \, \underbrace{e^{-y} \cdot \sin x}_{=:v(x,y)}$
From $\partial_x u(x,y) = - e^{-y} \cdot \cos(x) = \partial_y v(x,y) \\ \partial_y u(x,y) = - e^{-y} \cdot \cos(x) = - \partial_x v(x,y)$
we see that the Cauchy-Riemann equations are satisfied. Since the partial derivatives exist (and are continuous) we conclude that $z \mapsto e^{\imath \, z}$ is analytic. A similar argumentation shows that $z \mapsto e^{-\imath \, z}$ is analytic.