znanija.com/task/35212010
* * * * * * * * * * * * * * * * * * * * * * * * *
sin( arctg(1/2) - arcctg(-√3) )
Решение : arctg(1/2) = α , arcctg(-√3) =β
* * * определение: - π/2 < arctga < π/2 ; tg(arctga) = a * * *
arctg(1/2) ⇒ tgα = 1/2 ; 0 ≤ α < π/2
cosα = 1/ √(1+tg²α) = 1/√(1+(1/2)²) = 2/√5
sinα = tgα*cosα = (1/2)* (2/√5) = 1 / √5
arcctg(-√3) ⇒ tgβ = - √3 - π/2 < α < 0
cosβ = 1/ √(1+tg²β) = 1/√(1+(-√3)²) = 1/√2
sinβ = tgβ*cosβ = (-√3)*1/2 = - √3 /2 . * * * β = - π/3 = -60° * * *
sin( arctg(1/2) - arcctg(-√3) ) =
sin(arctg(1/2))*cos(arcctg(-√3)) - cos(sin( arctg(1/2) )*sin(arcctg(-√3)) =
sin(arcsin(1/√5 ))*cos(arccos(1/2))-cos(arccos(2/√5))*sin (arcsin(-√3/2)) =
( 1 /√5 )*(1/2) -(2/√5)*(-√3/2) = 1 /2√5 +√3 /√5 = (1 +2√3) /2√5 =
= √5 (1 +2√3) / 10 .