tg(2B+3пи/2), если sinB=-0,8

tg(2β+3пи/2) = ? Sinβ= - 0,8Решениеtg(2β + 3π/2) = - Ctg2β = -(1 - tg²β)/2tgβ tgβ = ?tgβ = Sinβ/CosβCos²β = 1 - Sin²β = 1 - 0,64 = 0,36, ⇒ Cosβ = 0,6tgβ = -0,8/0,6 = - 4/3tg(2β + 3π/2) = - Ctg2β = -(1 - tg²β)/2tgβ  = -(1+4/3)/2*(-4/3) = -7/3:(-8/3)==7/8