ВЗ
Валерий Зимин
Ответ. cos(2*a)=2*(cos(a))^2-1; (tg(a))^2=(1-(cos(a))^2)/((cos(a))^2)=1/((cos(a))^2)-1; (cos(a))^2=1/(1+(tg(a))^2:); cos(2*a)=2*(1/(1+(tg(a))^2))-1; cos(2*a)=2*(1/(1+(1/9))-1=0,8; ;
cos(a))^2=1/((tg(a))^2+1)=1/((-1/3)^2+1)=1/(7/6)=6/7,
cos(2*a)=(cos(a))^2-(sin(a))^2=2*(cos(a))^2-1=2*(6/7)-1=12/7-1=5/7
cos2α=(1-tg²α)/(1+tg²α)=(8/9)/(10/9)=4/5