Let α,β be the roots of x2−2xcos⁡ϕ+1=0  then the equation whose roots are αn and βn is

The given equation is  x2−2xcos⁡ϕ+1=0∴ x=2cos⁡ϕ±4cos2⁡ϕ−42=cos⁡ϕ±isin⁡ϕ Let α=cos⁡ϕ+isin⁡ϕ, then β=cos⁡ϕ−isin⁡ϕ ∴ αn+βn=(cos⁡ϕ+isin⁡ϕ)n+(cos⁡ϕ−isin⁡ϕ)n           =2cos⁡nϕ  and αnβn=(cos⁡nϕ+isin⁡nϕ)(cos⁡nϕ−isin⁡nϕ)=cos2⁡nϕ+sin2⁡nϕ=1∴ Required equation isx2−2xcos⁡nϕ+1=0