Given vector fields $\mathbf E$ and $\mathbf H$ with curl $\mathbf E= - \frac 1 c \frac {\partial \mathbf H} {\partial t}$ and curl $\mathbf H= \frac 1 c \frac {\partial \mathbf E} {\partial t}$ where $c$ is a constant, show the following:
$$\nabla \times (\nabla \times \mathbf E ) = - \frac 1 {c^2} \frac {\partial ^2 \mathbf E} {\partial t^2}$$
Thoughts: This is the curl of the curl of $\mathbf E$, which is the same as saying "curl $(-\frac 1 c \frac {\partial \mathbf H} {\partial t})$". I get the feeling I'm supposed to somehow substitute curl $\mathbf H$ for $\mathbf H$, allowing me to get rid of the curl on the outside of the parentheses. But I'm not sure if/how/why we're allowed to do that- the only reason I'd even suggest this is because the answer seems to fit. Help?
Assuming homogeneous&sourcleless medium, from maxwell equations we have:
$$\nabla \times (\nabla \times \mathbf E ) =\nabla \times (-\frac{\partial(\mu \mathbf H)}{\partial t})=-\mu\frac{\partial}{\partial t}\nabla \times \mathbf H=-\mu\frac{\partial}{\partial t}\frac{\partial(\epsilon \mathbf E)}{\partial t}=-\mu \epsilon\frac{\partial^2 \mathbf E}{\partial t^2}= - \frac 1 {c^2} \frac {\partial ^2 \mathbf E} {\partial t^2}$$