$(1+\cos \theta -i\sin \theta )(1+\cos 2\theta +i\sin 2\theta )=x+iy$  then  $x^2+y^2=\_\_\_\_\_\_\_\_$

$$\begin{gathered} (2{\cos }^2\frac{\theta }{2}-2i\sin \frac{\theta }{2}\cos \frac{\theta }{2})(2{\cos }^2\theta +i.2\sin \theta \cos \theta )=x+iy \\ \Rightarrow 4\cos \frac{\theta }{2}.\cos \theta cis\left(\frac{-\theta }{2}\right).cis\theta =x+iy \\ \Rightarrow 4\cos \frac{\theta }{2}\cos \theta cis\frac{\theta }{2}=x+iy \\ \therefore x=4{\cos }^2\frac{\theta }{2}\cos \theta ,y=4\cos \frac{\theta }{2}\sin \frac{\theta }{2}\cos \theta \\ {x}^2+y^2=16{\cos }^2\theta {\cos }^2\frac{\theta }{2} \end{gathered}$$