# In Solovay-Kitaev's algorithm, where does the rotation relation $\sin(\theta / 2) = 2 \sin^2(\phi/2)\sqrt{1 - \sin^4(\phi/2)}$ come from?

In Dawson's and Nielsen's pedagogical review of the Solovay-Kitaev algorithm, they describe the decomposition of U into $$U=VWV^\dagger W^\dagger$$, with both $$V, W$$ being unitary, being rotated by $$\phi$$ around the x-axis and y-axis, and obeying the following relationship, where $$\theta$$ is the angle of $$U$$ about the Bloch sphere:

$$\sin(\theta / 2) = 2 \sin^2(\phi/2)\sqrt{1 - \sin^4(\phi/2)}$$

Where does this relation come from?

This simply comes from equation an arbitrary rotation $$R_{n}(\theta)$$ with the rotation $$U=R_X(\phi)R_Y(\phi)R_X(-\phi)R_Y(-\phi).$$
The way that I did this calculation, just to verify this claim, was to recognise that, for example $$R_X(\phi)=\cos\frac{\phi}{2}I+i\sin\frac{\phi}{2}X.$$ Now, if I evaluate $$\text{Tr}(U)$$, this picks out the identity terms in the product (because I certainly don't want to multiply the whole thing out!), and those terms occur whenever I have a pair of $$X$$ or a pair of $$Y$$ (or both or neither). Hence \begin{align} \cos\frac{\theta}{2}&=\cos^4\frac{\phi}{2}+2\cos^2\frac{\phi}{2}\sin^2\frac{\phi}{2}-\sin^4\frac{\phi}{2} \\ &=(\cos^2\frac{\phi}{2}+\sin^2\frac{\phi}{2})^2-2\sin^4\frac{\phi}{2} \\ &=1-2\sin^4\frac{\phi}{2}. \end{align} Now if you use this to express $$\sin\frac{\theta}{2}$$, you'll find the claimed result.
• Just wanted to clarify, for the last step, one has to factor out $(\cos^2\theta/2 + \sin^2\theta/2)$ Commented Mar 17, 2021 at 23:15