已知α,β属于四分之三π,π,sin(α+β)=-五分之三,sin(β-四分之π)=十三分之十二,则cos(α+四分之π)
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数学之美团为你解答
α,β∈(3π/4,π)
α+β∈(3π/2,2π),β - π/4 ∈(π/2,3π/4)
sin(α+β) = - 3/5,sin(β - π/4) = 12/13
∴cos(α+β) = 4/5,cos(β - π/4) = - 5/13
cos(α+π/4) = cos [ (α+β) - (β - π/4) ]
= cos(α+β)cos(β - π/4) + sin(α+β)sin(β - π/4)
= 4/5 × (-5/13) + (-3/5) × 12/13
= - 56/65
数学之美团为你解答
α,β∈(3π/4,π)
α+β∈(3π/2,2π),β - π/4 ∈(π/2,3π/4)
sin(α+β) = - 3/5,sin(β - π/4) = 12/13
∴cos(α+β) = 4/5,cos(β - π/4) = - 5/13
cos(α+π/4) = cos [ (α+β) - (β - π/4) ]
= cos(α+β)cos(β - π/4) + sin(α+β)sin(β - π/4)
= 4/5 × (-5/13) + (-3/5) × 12/13
= - 56/65
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