1) cos(2arccos1\4)
2) sin(2arccos1\3)
В нашем случае cos(2arccos(1/4)=2*1/16-1=-7/8
2) sin(2arccos1/3)
Пусть arccos1/3= ∝ .0 ≤ ∝ ≤ π . тогда
cos ∝ =1/3; sin ∝ =sqrt(1-1/9)=sqrt(8/9).
sin(2arccos1/3=sin2 ∝ =2sin ∝ cos ∝ =2*2sqrt(2)/3*1/3=4sqrt(*2)/9