# In quantum search algorithm, how to interpret the effect of $U(t)$ as a rotation on the Bloch sphere?

In Nielsen's QCQI, in page 259, it reads,

$$U \left ( \Delta t \right ) = \left ( \cos^2 \left ( \frac {\Delta t} 2 \right ) - \sin ^2 \left ( \frac {\Delta t} 2 \right ) \vec \psi \cdot \hat z \right ) I$$ $$-2 i \sin \left ( \frac {\Delta t} 2 \right ) \left ( \cos \left ( \frac {\Delta t} 2 \right ) \frac {\vec \psi + \hat z} 2 + \sin \left ( \frac {\Delta t} 2 \right ) \frac {\vec \psi \times \hat z} 2 \right ) \cdot \vec \sigma$$ where $$U \left ( \Delta t \right )$$ is a operation of a Hamiltonian, $$\Delta t$$ is the time interval, $$\vec \psi$$ is the initial state.

Well, it seems complicated. But with $$\vec r = \cos \left ( \frac {\Delta t} 2 \right ) \frac {\vec \psi + \hat z} 2 + \sin \left ( \frac {\Delta t} 2 \right ) \frac {\vec \psi \times \hat z} 2$$ and $$\vec \psi \cdot \hat z = \frac 2 N -1$$ where $$N$$ is the number of the elements in the search space, it would be simplified to $$U \left ( \Delta t \right ) = \left (1-\frac 2 N \sin^2 \left ( \frac {\Delta t} 2 \right ) \right ) I -2 i \sin \left ( \frac {\Delta t} 2 \right ) \vec r \cdot \vec \sigma$$.

Then the book reads, $$U \left ( \Delta t \right )$$ is a rotation on the Bloch sphere about an axis of rotation $$\vec r$$ and through an angle $$\theta$$ defined by $$\cos \left ( \frac {\theta} 2 \right ) = 1-\frac 2 N \sin^2 \left ( \frac {\Delta t} 2 \right )$$.

My problem is, the definition of the rotation by $$\theta$$ about any $$\hat n$$ axis is $$R_{\hat n} \left ( \theta \right ) = \cos \left ( \frac \theta 2 \right ) I - i \sin \left ( \frac \theta 2 \right ) \hat n \cdot \vec \theta$$. Then in this case, $$\sin \left ( \frac \theta 2 \right ) = 2 \sin \left ( \frac {\Delta t} 2 \right )$$.

Then $$\sin^2 \left ( \frac \theta 2 \right ) + \cos^2 \left ( \frac \theta 2 \right ) \neq 1$$.

Where have I made a mistake?