Răspuns:
Vom avea nevoie de cosα si cosβ.
sinα=1/2 ; sin²α+cos²α= 1 => cos²α=1-(1/2)²= 1-1/4 => cos²α=3/4 => cosα= ±√3/2
Dar α ∈ ( 0, π/2) => cosα>0 => cosα= √3/2
sinβ=2/3 ; sin²β+cos²β=1 => cos²β=1-(2/3)²= 1-4/9 => cos²β=5/9 => cosβ=±√5/3
Dar β∈(0,π/2) => cosβ>0 => cosβ=√5/3
sin2α= 2sinαcosα
sin2a= 2*1/2*√3/2= √3/2
cos2α= 1-2sin²α= 1-2*(1/2)² = 1-2 * 1/4= 1-1/2=1/2
sin(α+β)= sinαcosβ + sinβcosα
= 1/2*√5/3 + 2/3*√3/2=√5/6 + √3/3 = 5+2√3/6
cos ( α-β)= cosαcosβ + sinαsinβ
=√3/2*√5/3 + 1/2*2/3=√15/6+1/3 = √15+2/6
tg(α+β)= tgα+tgβ/ 1-tgα*tgβ
unde tgα= sinα/cosα= 1/2 / √3/2= √3/3
tgβ= sinβ/cosβ= 2/3 / √5/3= 2√5/5
deci, tg(α+β)= √3/3+2√5/5 / 1- √3/3*2√5/5= (5√3+6√5)/15 / 1- 2√15/15
= (5√3+6√5)/15 / (15-2√15)/15 = 5√3+6√5/15-2√15 ( poti amplifica cu conjugata)
tg(α-β)= tgα - tgβ/ 1+ tgα*tgβ = √3/3 -2√5/5 / 1+√3/3*2√5/5 = (5√3-2√5)/15 / (15+2√15)/ 15 = (5√3-2√5)/15+2√15
Sper că te-am ajutat!
