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In the book Quarternion and Rotation Sequences, I can't seem to work out how the final equation (colored in $\color{red}{red}$) is derived from the original equation (colored in $\color{blue}{blue}$).
I did check using Wikipedia's list of trigonometric identities as my references as well as the book errata but to no avail.
I copy/paste the text that is giving me problem literally below:

Thus our tracking transformation has axis of rotation given by $$\color{blue}{v=\left(k,\frac{k\sin\alpha}{\cos\alpha-1},\frac{k\sin\beta}{\cos\beta-1}\right).}$$
Notice that in this computation we determine only the direction of the axis of rotation.
Should we wish to obtain a specific vector as the axis of rotation we may,
for instance, choose $k = -1$, to obtain
We note that by using the trigonometric identity
we may write the following expression for the axis of the rotation $$\color{red}{v=\left(-\sin\frac{\alpha}{2}\sin\frac{\beta}{2},\cos\frac{\alpha}{2}\sin\frac{\beta}{2},\sin\frac{\alpha}{2}\cos\frac{\beta}{2}\right)}$$

Anyone have any idea?

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up vote 3 down vote accepted


  1. $1-\cos{2\theta}=2\sin^2\frac{\theta}{2}$.
  2. $\sin{\theta}=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$, which we can rearrange to $\displaystyle\frac{\sin{\theta}}{2\sin\frac{\theta}{2}}=\cos\frac{\theta}{2}$.


So, starting with $$\color{blue}{\vec{v}=\left(k,\frac{k\sin\alpha}{\cos\alpha-1},\frac{k\sin\beta}{\cos\beta-1}\right)}.$$

Since it doesn't matter what $k$ we use, pick $k=-1$ and this becomes

$$\vec{v}=\left(-1,\frac{\sin{\alpha}}{1-\cos{\alpha}},\frac{\sin{\beta}}{1-\cos{\beta}}\right).$$ We can use fact 1 to replace the denominators of the second and third components, obtaining $$\vec{v}=\left(-1,\frac{\sin{\alpha}}{2\sin^2\frac{\alpha}{2}},\frac{\sin{\beta}}{2\sin^2\frac{\beta}{2}}\right).$$ Now multiply through by $\sin\frac{\alpha}{2}\sin\frac{\beta}{2}$ and we get $$\vec{v}=\left(-\sin\frac{\alpha}{2}\sin\frac{\beta}{2},\frac{\sin\alpha\sin\frac{\beta}{2}}{2\sin\frac{\alpha}{2}},\frac{\sin{\beta}\sin\frac{\alpha}{2}}{2\sin\frac{\beta}{2}}\right).$$ Finally, we use fact 2 on the second and third terms. $$\color{red}{\vec{v}=\left(-\sin\frac{\alpha}{2}\sin\frac{\beta}{2},\cos\frac{\alpha}{2}\sin\frac{\beta}{2},\cos\frac{\beta}{2}\sin\frac{\alpha}{2}\right)}.$$

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thanks, I didn't know that you could just multiply the equation by $\sin\frac{\alpha}{2}\sin\frac{\beta}{2}$ on Right Hand Side only. – kypronite Oct 7 '12 at 5:42
Yeah - since it is a vector, doing so only affects the magnitude, and this proof only aims to find the direction of $\vec{v}$. – Alexander Gruber Oct 7 '12 at 5:48
ic,yeah.That made much more sense. – kypronite Oct 7 '12 at 10:41

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