Cos 2альфа ,если tg(альфа-2п/3)=2-корень из3

Tg(a-2π/3)=(tga-tg2π/3)/(1+tga*tg2π/3)=(tga+√3/2)/(1-√3/2tga)=2-√3
tga+√3/2=(1-√3/2tga)*(2-√3)
tga+√3/2=2-√3-√3tga+3/2
tga+√3tga=2-√3+3/2-√3/2
tga(1+√3)=(7/2-3√3/2)
tga=(7/2-3√3/2)/(1+√3)=(7/2-3√3/2)*(√3-1))/(3-1)=(7/4-3√3/4)(√3-1)=
=7√3/4-7/4-9/4+3√3/4)=10√3/4-4=2,5√3-4
cos²a=1:(1+tg²a)=1:(1+18,75-20√3+16)=1/(35,75-20√3)
cos2a=2cos²a-1=2/(35,75-20√3) - 1=(2-35,75+20√3)/(35,75-20√3=
=(-33,75+20√3)/(35,75-20√3)