Sin(arctg1/2-arcctg(-√3))
С объяснением

* * * * * * * * * * * * * * * * * * * * * * * * *

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 .