найдите
cos(2α+β), если ctgβ= -(12/5), cos 2α= 1/4, π<2β<2π, 3π/4<α<π

найдитеcos(2α+β), если ctgβ= -(12/5), cos 2α= 1/4, π<2β<2π, 3π/4<α<π-------cos(2α+β) =cos2α*cosβ -sin2α*sinβ;* * *3π/4<α<π  ⇔ 3π/2 <2α<2π ⇒sin2α <0 * * *sin2α = -√(1-cos²2α) = -√(1-(1/4)²)= -(√15)/4 .  * * *π<2β<2π  ⇔ π/2<β<π  ⇒sinβ >0 * * *sinβ = 1/√(1+ctq²β)=1/√(1+144/25) =5/13.cosβ =sinβ*ctqβ = (5/13)*(-12/5) = -12/13.---cos(2α+β) =cos2α*cosβ -sin2α*sinβ =(1/4)*(-12/13) +(√15)/4)*(5/13)=-12/52 +5√15/52 =(5√15 -12)/52.