求详解,一步步来,对于我这种数学渣渣
1个回答

由sinβ=-12/13且β∈(-π/2,0),得cosβ = 5/13

由α∈(π/2,π),β∈(-π/2,0),得2α-β∈(π,5π/2)

又sin(2α-β) = 3/5 > 0,所以2α-β∈(2π,5π/2)

sin(2α-β) = sin(2α)cosβ - cos(2α)sinβ = 5/13sin(2α) + 12/13cos(2α) = 3/5

cos(2α-β) = cos(2α)cosβ + sin(2α)sinβ = 5/13cos(2α) - 12/13sin(2α) = 4/5

解得cos(2α) = 56/65

sin²α = (1/2)(1 - cos(2α)) = 9 / 130

sinα = 3/√130

f(x)=(2cos²x-1)-6cosx+1

=2cos²x-6cosx

=2(cosx-3/2)²-9/2

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