Question

已知α,β∈（0,π,且tanα=2,cosβ=-7√2／10.求cos2α,2α-β的值

Answer 1

由万能公式,cos2α = (1 –tan2α)/(1 + tan2α) = (1 – 22)/(1 + 22)= -3/5 ；
sin2α =(2tanα)/(1 + tan2α) = 2*2/(1 + 22) = 4/5 ；
因为cosβ = -7√2/10 0,β∈(0,π),所以β在第二象限,β∈(π/2,π),sinβ > 0,所以sinβ = √(1 – cos2β) = √(1 – 49*2/100) = √2/10 .
所以cos(2α – β)= cos2αcosβ + sin2αsinβ = (-3/5)(-7√2/10) + (4/5)(√2/10) = 25√2/50 = √2/2 ,而且sin(2α – β) = sin2αcosβ – cos2αsinβ = (4/5)(-7√2/10) – (-3/5)(√2/10) = -25√2/50 = -√2/2 ,所以2α – β = arcsin(-√2/2) = -π/4 ；
综上所述,cos2α = -3/5,2α – β = -π/4 .