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

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?

• Why do not the inverse to shift onto the right side and to make some trigonometric transformations ? Maybe these formulas can help en.wikipedia.org/wiki/… – Widawensen Mar 5 '19 at 12:34
• Additionally we also have $\hat{\omega}^3=-\hat{\omega}$ – Widawensen Mar 5 '19 at 12:41
• @Widawensen Thanks for your enlightening implications. I have finished the proof on my scratch paper! :) – Finley Mar 5 '19 at 12:55
• I'm glad to be helpful :) – Widawensen Mar 5 '19 at 12:56

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}
{"ft_lang_label":"__label__en","ft_lang_prob":0.5474531,"math_prob":0.999936,"size":2243,"snap":"2020-10-2020-16","text_gpt3_token_len":823,"char_repetition_ratio":0.23581956,"word_repetition_ratio":0.11221122,"special_character_ratio":0.40035665,"punctuation_ratio":0.056930695,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999939,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-03T21:16:26Z\",\"WARC-Record-ID\":\"<urn:uuid:32d3c990-4dfe-4151-bc49-071369860a3f>\",\"Content-Length\":\"146810\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:60a85a4a-5706-4f6f-8f4b-16e9b0118776>\",\"WARC-Concurrent-To\":\"<urn:uuid:40976daf-25e8-4cef-ae97-de3582016cad>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3136142/solve-for-the-inverse-of-mathbf-i-tan-frac-phi2-mathbf-hat-omega\",\"WARC-Payload-Digest\":\"sha1:WGDGM6PX5XMLW2YNDEPFFOHP3PT5TQZN\",\"WARC-Block-Digest\":\"sha1:ILOO67DYPWFOYRAGWTMHXSH7DODUUVMD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370518622.65_warc_CC-MAIN-20200403190006-20200403220006-00107.warc.gz\"}"}