# Quaternions - Prove that two quaternions map to the same R

I am given the following question: I need to prove that two quaternions map to the same Rotation Matrix in SO3 Space. It is demonstrated by this image:

Let w be v here. I tried to work out the proof, but it isn't coming out correctly:

$$Q=(cos\frac{\theta}{2},w.sin\frac{\theta}{2})$$

$$Q=cos\frac{\theta}{2}+w_{1}.sin\frac{\theta}{2}.\mathbf{i}+w_{2}.sin\frac{\theta}{2}.\mathbf{j}+w_{3}.sin\frac{\theta}{2}.\mathbf{k}$$

As per image, replace $$\theta$$ with $$2\pi-\theta$$ and w with -w

$$Q=cos\frac{2\pi-\theta}{2}+-w_{1}.sin\frac{2\pi-\theta}{2}.\mathbf{i}+-w_{2}.sin\frac{2\pi-\theta}{2}.\mathbf{j}+-w_{3}.sin\frac{2\pi-\theta}{2}.\mathbf{k}$$

$$Q=-cos\frac{\theta}{2}+w_{1}.sin\frac{\theta}{2}.\mathbf{i}+w_{2}.sin\frac{\theta}{2}.\mathbf{j}+w_{3}.sin\frac{\theta}{2}.\mathbf{k}$$

It is not equal! I am left with $$-cos\frac{\theta}{2}$$ which is not matching. Please tell me how to fix this

• You're told that there are two ways to encode the rotation, and you're trying to prove the two ways are one. What did you expect :) – rschwieb Jan 28 at 11:53
• – Jyrki Lahtonen Jan 31 at 18:48

Since $$\sin(\pi-\theta/2)$$ equals $$\sin(\theta/2)$$, not $$-\sin(\theta/2)$$, you're missing minus signs.
Note $$\cos(\theta/2)+\sin(\theta/2)\mathbf{w}$$ is expressible as $$Q=\exp(\frac{1}{2}\theta\mathbf{w})$$ (assuming $$\|\mathbf{w}\|=1$$).
Replacing $$\theta\mapsto2\pi-\theta$$ and $$\mathbf{w}\mapsto -\mathbf{w}$$ yields $${\bf\color{Red}{-}}Q$$, not $$Q$$; you shouldn't be expecting the original quaternion in the first place. Both unit quaternions $$Q$$ and $$-Q$$ represent the same 3D rotation.
• Do you know how quaternions are used to represent 3D rotations? Given a quaternion $q$, with polar form $q=\exp(\frac{1}{2}\theta{\bf w})$, and a 3D vector $\bf x$ (in other words, a purely imaginary quaternion), the conjugation $q{\bf x}q^{-1}$ has the effect of rotating $\bf x$ around $\bf w$ by the angle $\theta$. Since $q{\bf x}q^{-1}=(-q){\bf x}(-q)^{-1}$ for all $\bf x$, we conclude $q$ and $-q$ represent the same 3D rotation. – arctic tern Jan 31 at 2:41