Rodrigues Formula

Question

So in my homework on rotation matrices, I stumbled upon this question:

The question asks for $\sin(\theta)$. So could any of you elaborate on how to approach the question. Note that $J$ is the skew symmetric matrix.

Answer 1

Here is the derivation for the trace relationship mentioned above from the Rodrigues formula:

$$ R = I + J(\hat{\omega}) \sin\theta + (1-cos\theta) J(\hat{\omega})^{2} $$

Say $\hat{\omega} = (\omega_{1}, \omega_{2}, \omega_{3})^{T}$ then $J(\hat{\omega})$ can be written

$$ J(\hat{\omega}) = \begin{pmatrix} 0 & -\omega_{3} & \omega_{2} \\ \omega_{3} & 0 & -\omega_{1} \\ -\omega_{2} & \omega_{1} & 0 \end{pmatrix} $$

$$ J^{2}(\hat{\omega}) =\begin{pmatrix} -\omega_{2}^{2} - \omega_{3}^{2} & \omega_{1}\omega_{2} & \omega_{1}\omega_{3} \\ \omega_{1}\omega_{2} & -\omega_{1}^{2} - \omega_{3}^{2} & \omega_{2}\omega_{3} \\ \omega_{1}\omega_{3} & \omega_{2}\omega_{3} & -\omega_{1}^{2} - \omega_{2}^{2} \end{pmatrix} $$

We obtain the trace of these matrices:

$$ \textrm{trace} (J(\hat{\omega}) ) = 0 $$

$$ \textrm{trace}( J^{2}(\hat{\omega})) = -2 (\omega_{1}^{2} + \omega_{2}^{2} + \omega_{3}^{2}) = -2 $$

Putting this together gives

$$ \begin{aligned} \textrm{trace} (R) &= \textrm{trace} (I) + \sin\theta \, \textrm{trace} (J(\hat{\omega}) ) + (1-cos\theta) \, \textrm{trace}(J^{2}(\hat{\omega})) \\ &= 3 - 2 (1 - \cos\theta) \\ &= 1 + 2\cos\theta \end{aligned} $$

So, in summary, if $\hat{\omega}$ is known, then $\theta$ can be calculated from the trace of $R$.

Answer 2

If one has a $3$-by-$3$ rotation matrix through an angle $\theta$, then its trace $t$ (the sum of its diagonal entries) equal $1+2\cos\theta$. So $\cos\theta=\frac12(t-1)$, and then $\sin\theta=\sqrt{1-\cos^2\theta}$.