If $\left(1+\cos \mathrm{\theta }+\mathrm{i} \sin \mathrm{\theta }\right) \left(1+\cos 2\mathrm{\theta }+\mathrm{i} \sin 2\mathrm{\theta }\right)=\mathrm{x}+\mathrm{iy}$ then $\mathrm{y}=$

$\mathrm{x}+\mathrm{iy}=\left(1+\cos \mathrm{\theta }+\mathrm{i} \sin \mathrm{\theta }\right) \left(1+\cos 2\mathrm{\theta }+\mathrm{i} \sin 2\mathrm{\theta }\right)$

$\begin{array}{rcl}\mathrm{x}+\mathrm{iy}& =& \left(2{\cos }^2\frac{\mathrm{\theta }}{2}+\mathrm{i} 2\sin \frac{\mathrm{\theta }}{2} \cos \frac{\mathrm{\theta }}{2}\right) \left(2{\cos }^2\mathrm{\theta }+\mathrm{i} 2\mathrm{sin\theta } \mathrm{cos\theta }\right)\end{array}$

$\begin{array}{rcl}\mathrm{x}+\mathrm{iy}& =& 2\cos \frac{\mathrm{\theta }}{2}·2\mathrm{cos\theta } \left(\cos \frac{\mathrm{\theta }}{2}+\mathrm{i} \sin \frac{\mathrm{\theta }}{2}\right) \left(\mathrm{cos\theta }+\mathrm{i} \mathrm{sin\theta }\right)\end{array}$

$\begin{array}{rcl}\mathrm{x}+\mathrm{i} \mathrm{y}& =& 2^2 \cos \frac{\mathrm{\theta }}{2} \mathrm{cos\theta } \mathrm{cis}\left(\frac{\mathrm{\theta }}{2}+\mathrm{\theta }\right)\end{array}$

$\begin{array}{rcl}\mathrm{x}+\mathrm{i} \mathrm{y}& =& 4\cos \frac{\mathrm{\theta }}{2} \mathrm{cos\theta }\left(\cos \frac{3\mathrm{\theta }}{2}+\mathrm{i} \sin \frac{3\mathrm{\theta }}{2}\right)\end{array}$

$\begin{array}{rcl}\frac{\mathrm{y}}{\mathrm{x}}& =& \frac{\sin 3 \mathrm{\theta }/2}{\cos {\displaystyle \frac{3\mathrm{\theta }}{2}}} \Rightarrow \mathrm{y}=\tan \frac{3\mathrm{\theta }}{2} \mathrm{x}\end{array}$