Solve for the inverse of $\mathbf I - \tan(\frac{\phi}{2}) \mathbf {\hat \omega}$

Question

Original problem comes from some notes on rotations (at the last page), which was devoted to deriving Rodrigues' rotation formula. The complete problem is to show why $$(\mathbf I - \tan(\frac{\phi}{2}) \mathbf {\hat \omega})^{-1}(\mathbf I + \tan(\frac{\phi}{2}) \mathbf {\hat \omega}) = \mathbf I + \sin \phi \mathbf {\hat \omega} +(1 - \cos \phi)\mathbf {\hat \omega}^2.$$ Here, $\mathbf {\hat \omega}$ is a skew-symmetric matrix: $$\mathbf {\hat \omega} = \begin{bmatrix} 0 & - \omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{bmatrix}$$ I think the key is to solve for the inverse of $\mathbf I - \tan(\frac{\phi}{2}) \mathbf {\hat \omega}$, can anyone help out?

Answer 1

It boils down to show: \begin{equation} \begin{aligned} (\mathbf I + \tan(\frac{\phi}{2}) \mathbf {\hat \omega}) &=(\mathbf I - \tan(\frac{\phi}{2}) \mathbf {\hat \omega})( \mathbf I + \sin \phi \mathbf {\hat \omega} +(1 - \cos \phi)\mathbf {\hat \omega}^2) \\ &=\mathbf I + \sin \phi{\hat \omega} + (1-\cos \phi){\hat \omega}^{2}-\tan \frac{\phi}{2} {\hat \omega} - \sin \phi \tan \frac{\phi}{2}{\hat \omega}^2+\tan \frac{\phi}{2} (1-\cos\phi){\hat \omega} \\ &= \mathbf I + (\sin \phi-\tan \frac{\phi}{2}\cos \phi){\hat \omega}+(1-\cos\phi-\tan \frac{\phi}{2}\sin \phi){\hat \omega}^2. \end{aligned} \end{equation} Now let $t=\tan \frac{\phi}{2}$, and $\cos \phi=\frac{1-t^2}{1+t^2},\sin \phi=\frac{2t}{1+t^2}$. So above expression is equal to \begin{equation} \begin{aligned} & \mathbf I + (\frac{2t}{1+t^2}-\frac{t-t^3}{1+t^2}){\hat \omega}+ (1-\frac{1-t^2}{1+t^2}-\frac{2t^2}{1+t^2}){\hat \omega}^2 \\ =& \mathbf I + t\hat \omega \end{aligned} \end{equation}