Characteristic function - how to guess distribution

Question

I have random variable $Z$ with characteristic function of the form $\cos^3(t)$. How to calculate characteristic function of $Y=Z^3+2Z^2+1$

Answer 1

Since $Z$ has characteristic function$$((e^{it}+e^{-it})/2)^3=\tfrac18(e^{3it}+3e^{it}+3e^{-it}+e^{-3it}),$$$P(Z=1)=P(Z=-1)=\frac38$ while $P(Z=3)=P(Z=-3)=\frac38$, so$$P(Y=4)=P(Y=2)=\tfrac38,\,P(Y=46)=P(Y=-8)=\tfrac18.$$Hence $Y$ has characteristic function $\frac18(e^{-8it}+3e^{2it}+3e^{4it}+e^{46it})$.