设2sinα=sinθ+cosθ①,sin2β=sinθcosθ②
由①2-2×②得4sin2α-2sin2β=1即2(1-cos2α)-(1-cos2β)=1,2cos2α=cos2β
∴cos4β-4cos4α=2cos22β-1-4(2cos22α-1)
=2(2cos2α)2-8cos22α+3=3
设2sinα=sinθ+cosθ①,sin2β=sinθcosθ②
由①2-2×②得4sin2α-2sin2β=1即2(1-cos2α)-(1-cos2β)=1,2cos2α=cos2β
∴cos4β-4cos4α=2cos22β-1-4(2cos22α-1)
=2(2cos2α)2-8cos22α+3=3