Definition. The dual 1-forms $\theta_1,\theta_2,\theta_3$ of a frame $E_1,E_2,E_3$ on $\mathbb{E}^3$ are defined by
$$\theta_i(v)=v\cdot E_i(p),\ v\in T_p\mathbb{E}^3.$$
Clearly $\theta_i$ is linear.
Example. The dual 1-forms of the natural frame $U_1,U_2,U_3$ are $dx_1$, $dx_2$, $dx_3$ since
$$dx_i(v)=v_i=v\cdot U_i(p)$$
for each $v\in T_p\mathbb{E}^3$.
For any vector field $V$ on $\mathbb{E}^3$,
$$V=\sum_i\theta_i(V)E_i.$$
To see this, let us calculate for each $V(p)\in T_p\mathbb{E}^3$
\begin{align*}
\sum_i\theta_i(V(p))E_i(p)&=\sum_i(V(p)\cdot E_i(p))E_i(p)\\
&=\sum_iV_i(p)E_i(p)\\
&=V(p).
\end{align*}
Lemma. Let $\theta_1,\theta_2,\theta_3$ be the dual 1-forms of a frame $E_1, E_2, E_3$. Then any 1-form $\phi$ on $\mathbb{E}^3$ has a unique expression
$$\phi=\sum_i\phi(E_i)\theta_i.$$
Proof. Let $V$ be any vector field on $\mathbb{E}^3$. Then
\begin{align*}
\sum_i\phi(E_i)\theta_i(V)&=\sum_i\phi(E_i)\theta_i(V)\\
&=\phi(\sum_i\theta_i(V)E_i)\ \mbox{by linearity of $phi$}\\
&=\phi(V).
\end{align*}
Let $A=(a_{ij})$ be the attitude matrix of a frame field $E_1$, $E_2$, $E_3$, i.e.
\begin{equation}\label{eq:frame}E_i=\sum_ja_{ij}U_j,\ i=1,2,3.\end{equation}
Clearly $\theta_i=\sum_j\theta_i(U_j)dx_j$. On the other hand,
$$\theta_i(U_j)=E_i\cdot U_j=\left(\sum_ka_{ik}U_k\right)\cdot U_j=a_{ij}.$$ Hence the dual formulation of \eqref{eq:frame} is
\begin{equation}\label{eq:dualframe}\theta_i=\sum_ja_{ij}dx_j.\end{equation}
Theorem. [Cartan Structural Equations] Let $E_1$, $E_2$, $E_3$ be a frame field on $\mathbb{E}^3$ with dual 1-forms $\theta_1$, $\theta_2$, $\theta_3$ and connection forms $\omega_{ij}$, $i,j=1,2,3$. Then
- The First Structural Equations: $$d\theta_i=\sum_j\omega_{ij}\wedge\theta_j.$$
- The Second Structural Equations: $$d\omega_{ij}=\sum_k\omega_{ik}\wedge\omega_{kj}.$$
Proof. The exterior derivative of \eqref{eq:dualframe} is
$$d\theta_i=\sum_jda_{ij}\wedge dx_j.$$ Since $\omega=dA\cdot{}^tA$ and ${}^tA=A^{-1}$ (recall that $A$ is an orthogonal matrix), $dA=\omega\cdot A$, i.e.
$$da_{ij}=\sum_k\omega_{ik}a_{kj}.$$
So,
\begin{align*}
d\theta_i&=\sum_j\left\{\left(\sum_k\omega_{ik}a_{kj}\right)\wedge dx_j\right\}\\
&=\sum_k\left\{\omega_{ik}\wedge\sum_j a_{kj}dx_j\right\}\\
&=\sum_k\omega_{ik}\wedge\theta_k.
\end{align*}
From $\omega=dA\cdot{}^tA$,
\begin{equation}\label{eq:connectform}\omega_{ij}=\sum_kda_{ik}a_{jk}.\end{equation}
The exterior derivative of \eqref{eq:connectform} is
\begin{align*}
d\omega_{ij}&=\sum_k da_{jk}\wedge d_{ik}\\
&=-\sum_k da_{ik}\wedge da_{jk},
\end{align*}
i.e.
\begin{align*}
d\omega&=-dA\wedge{}^t(dA)\\
&=-(\omega\cdot A)\cdot({}^tA\cdot{}^t\omega)\\
&=-\omega\cdot (A\cdot{}^tA)\cdot{}^t\omega\\
&=-\omega\cdot{}^t\omega\ \ \ (A\cdot{}^tA=I)\\
&=\omega\cdot\omega.\ \ \ (\mbox{$\omega$ is skew-symmetric.})
\end{align*}
This is equivalent to the second structural equations.
Example. [Structural Equations for the Spherical Frame Field] Let us first calculate the dual forms and connection forms.
From the spherical coordinates
\begin{align*}
x_1&=\rho\cos\varphi\cos\theta,\\
x_2&=\rho\cos\varphi\sin\theta,\\
x_3&=\rho\sin\varphi,
\end{align*}
we obtain differentials
\begin{align*}
dx_1&=\cos\varphi\cos\theta d\rho-\rho\sin\varphi\cos\theta d\varphi-\rho\cos\varphi\sin\theta d\theta,\\
dx_2&=\cos\varphi\sin\theta d\rho-\rho\sin\varphi\sin\theta d\varphi+\rho\cos\varphi\cos\theta d\theta,\\
dx_3&=\sin\varphi d\rho+\rho\cos\varphi d\varphi.
\end{align*}
From the spherical frame field $F_1$, $F_2$, $F_3$ discussed here, we find its attitude matrix
$$A=\begin{pmatrix}
\cos\varphi\cos\theta & \cos\varphi\sin\theta & \sin\varphi\\
-\sin\theta & \cos\theta & 0\\
-\sin\varphi\cos\theta & -\sin\varphi\sin\theta & \cos\varphi
\end{pmatrix}.$$
Thus by (2) we find the dual 1-forms
\begin{align*}
\begin{pmatrix}
\theta_1\\
\theta_2\\
\theta_3
\end{pmatrix}&=\begin{pmatrix}
\cos\varphi\cos\theta & \cos\varphi\sin\theta & \sin\varphi\\
-\sin\theta & \cos\theta & 0\\
-\sin\varphi\cos\theta & -\sin\varphi\sin\theta & \cos\varphi
\end{pmatrix}\begin{pmatrix}
dx_1\\
dx_2\\
dx_3
\end{pmatrix}\\
&=\begin{pmatrix}
d\rho\\
\rho\cos\theta d\theta\\
\rho d\varphi
\end{pmatrix}.
\end{align*}
\begin{align*}
&dA=\\
&\begin{bmatrix}
-\sin\varphi\cos\theta d\varphi-\cos\varphi\sin\theta d\theta & -\sin\varphi\sin\theta d\varphi+\cos\varphi\cos\theta d\theta & \cos\varphi d\varphi\\
-\cos\theta d\theta & -\sin\theta d\theta & 0\\
-\cos\varphi\cos\theta d\varphi+\sin\varphi\sin\theta d\theta & -\cos\varphi\sin\theta d\varphi-\sin\varphi\sin\theta d\theta & -\sin\varphi d\varphi
\end{bmatrix}\end{align*}
and so,
\begin{align*}
\omega&=\begin{pmatrix}
0 & \omega_{12} & \omega_{13}\\
-\omega_{12} & 0 & \omega_{23}\\
-\omega_{13} & -\omega_{23} & 0
\end{pmatrix}\\
&=dA\cdot{}^tA\\
&=\begin{pmatrix}
0 & \cos\varphi d\theta & d\varphi\\
-\cos\varphi d\theta & 0 & \sin\varphi d\theta\\
-d\varphi & -\sin\varphi d\theta & 0
\end{pmatrix}.
\end{align*}
From these dual 1-forms and connections forms one can immediately verify the first and the second structural equations.