arcsin(sqrt(2/3))=α; arcsin(((√2)–1)/√6)=β
Найти
α-β
arcsin(sqrt(2/3))=α,⇒ sin α=sqrt(2/3)
и α∈[-π/2;π/2]
Найдем cosα=sqrt(1-sin^2α)=
=sqrt(1-(sqrt(2/3))^2)=sqrt(1/3).
arcsin(((√2)–1)/√6)=β ⇒ sinβ =((√2)–1)/√6
и β∈[-π/2;π/2]
Найдем cosβ=sqrt(1-sin^2β)=sqrt(1-(((√2)–1)/√6)^2)=sqrt(((√2)+1)/√6)
Найдем
sin(α-β)=sinαcosβ-cosαsinβ=
=(2+sqrt(2)-sqrt(1)-1)/sqrt(18)=1/sqrt(2)
α-β=π/4
О т в е т. π/4