Задача 33862 ...
Условие
1) arccos (7)/(8) = 2arcsin (1)/(4)
2)arcsin(4)/(5) + arcsin(5)/(13) + arcsin (16)/(65) = π/2
Решение
sin(arccos(7/8))=sin(2arcsin(1/4))
Считаем
sin(arccos(7/8))
Пусть arccos(7/8)= α , α ∈ [0;π/2]
cos α =7/8;
sin α =+sqrt(1-cos^2 α )=sqrt(1-(7/8)^2)=sqrt(15)/8
считаем
sin(2arcsin(1/4))
arcsin(1/4)= β , β ∈ [0;π/2]
sin β =1/4
cos β =sqrt(1-(1/4)^2)=sqrt(15)/4
sin2 β =2*sin β *cos β = 2*(1/4)*sqrt(15)/4
правая часть sqrt(15)/8 равна левой 2*(1/4)*sqrt(15)/4
верно.
2)
arcsin(4)/(5) + arcsin(5)/(13) + arcsin (16)/(65) = π/2
sin(arcsin(4)/(5) + arcsin(5)/(13) + arcsin (16)/(65))=sin(π/2)
arcsin(4/5)= α ⇒ sin α =4/5; α ∈ [0;π/2] ; cos α =3/5
arcsin(5)/(13)= β ⇒ sin β =5/13 ; β ∈ [0;π/2] ; cos β =12/13
arcsin (16)/(65)= γ ⇒ sinγ =16/65 ; γ ∈ [0;π/2] ; cos γ =63/65
sin( α + β + γ )=sin(( α + β )+ γ )=sin( α + β )*cos γ +cos( α + β )*sin γ =
=(sin α cos β +cos α sin β )*cos γ +(cos α *cos β -sin α *sin β )*sin γ
=((4/5)*(12/13)+(3/5)*(5/13)) * (63/65) + ((3/5)*(12/13) - (4/5)*(5/13)) * (16/65)=
=(63/65)^2+(16/65)^2=(3969+256)/4225=4225/4225=1
sin(π/2)=1
О т в е т. верно