Найдитеcos(2α+β), если ctgβ= -(12/5), cos 2α= 1/4, π<2β<2π, 3π/4<α<π-------<br>cos(2α+β) =cos2α*cosβ -sin2α*sinβ;
* * *3π/4<α<π ⇔ 3π/2 <2α<2π ⇒<span>sin2α <0</span> * * *
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.
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cos(2α+β) =cos2α*cosβ -sin2α*sinβ =(1/4)*(-12/13) +(√15)/4)*(5/13)=
-12/52 +5√15/52 =(5√15 -12)/52.