Quantum Computing 1.4 -- Bloch sphere

The angle $\theta$ can be derived from the z-coordinate of the vector. With the help of the cosine additon formula and the Pythagorean theorem, we get an expression for $\cos( \theta/2)$ and $\sin(\theta/2)$ : $\begin{aligned} & & z = \cos \theta &= \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} & & |\text{cosine addition formula} \\ & & &= 2 \cos^2 \frac{\theta}{2} - 1 & &|\text{Pythagoras } \cos^2 x + \sin^2 x = 1 \\ & & &= 1 - 2 \sin^2 \frac{\theta}{2} \\ \Rightarrow & & \cos \frac{\theta}{2}& = \sqrt{\frac{1 + \cos \theta}{2}} = \sqrt{\frac{1 + z}{2}}, \\ & & \sin \frac{\theta}{2}& = \sqrt{\frac{1 - \cos \theta}{2}} = \sqrt{\frac{1 - z}{2}} \end{aligned}$ The phase angle $\phi$ is determined by the ratio between y- and x-coordinate: $\begin{aligned} & & \frac{y}{x} &= \frac{\sin \phi}{\cos \phi}\\ & & &= \frac{\sqrt{1 - \cos^2 \phi}}{\cos \phi} & & |\text{Pythagoras } \cos^2 x + \sin^2 x = 1 \\ & & &= \sqrt{\frac{1}{\cos^2 \phi} - 1} \\ \Rightarrow & & \cos \phi &= \frac{1}{\sqrt{1 + (y/x)^2}}, \\ \Rightarrow & & \sin \phi &= \sqrt{1 - \cos^2 \phi} \\ & & &= \frac{y/x}{\sqrt{1 + (y/x)^2}} \\ \Rightarrow & & e^{i \phi} &= \cos \phi + i \sin \phi \\ & & &= \frac{1 + i y/x}{\sqrt{1 + (y/x)^2}}\\ \end{aligned}$ The general solution for the state is accordingly $|\psi\rangle = \sqrt{\frac{1 + z}{2}} |0\rangle + \sqrt{\frac{1 - z}{2}} \frac{1 + i y/x}{\sqrt{1 + (y/x)^2}} |1 \rangle$ in our case is $z = 1/2$ and $x=y = \sqrt{3/8}$ , so that $|\psi\rangle = \frac{\sqrt{3}}{2} |0\rangle + \frac{1 + i }{\sqrt{8}} |1 \rangle$