I believe that the sign of \sin(\theta/2) should be positive in Exercise I.6.4
Hey @AvanishM! Welcome to the forum 
The equations here are correct 
. The X operator flips the qubit state:
X \vert 0 \rangle = \vert 1 \rangle,
X \vert 1 \rangle = \vert 0 \rangle.
So, the \vert 1 \rangle state with the \sin(\theta/2) coefficient flips to \vert 0 \rangle. Let me know if this makes sense!
Yes that makes sense, but I was actually looking at the first step, where we substitute RY(\theta) \vert 0 \rangle!
RY(\theta) \vert 0 \rangle = \cos(\theta/2) \vert 0 \rangle + \sin(\theta/2) \vert 1 \rangle
This example seems to have substituted:
RY(\theta) \vert 0 \rangle = \cos(\theta/2) \vert 0 \rangle - \sin(\theta/2) \vert 1 \rangle
What’s in the codebook is correct
RY(\theta) \vert 0 \rangle = \begin{pmatrix} \cos(\theta/2) & -\sin(\theta/2) \\ \sin(\theta/2) & \cos(\theta/2) \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \cos(\theta/2) + 0 \\ \sin(\theta/2) + 0 \end{pmatrix} = \cos(\theta/2)\vert 0 \rangle + \sin(\theta/2)\vert 1 \rangle