sin4α-cos4α=2sin(2α)*cos(2α)-(1-2*sin(2α)^2)=2sin(2α)*cos(2α)+2*sin(2α)^2-1=2*sin(2α)(cos(2α)+sin(2α))-1cos(2α)+sin(2α)不恒于1 sin4α-cos4α=2sin2α-1 不是恒等式sinx/1-cosx=1+cosx/sinx 不...你錯(cuò)了 它們都是恒等式sin^4α-cos^4α=2sin^2α-1左式=(sin^2α+cos^2α)(sin^2α-cos^2α)=sin^2α-cos^2α=sin^2α-1+sin^2α=2sin^2α-1=右式sinx/(1-cosx)=(1+cosx)/sinx左式=sinx(1+cosx)/((1+cosx)(1-cosx))=sinx(1+cosx)/(1-cos^2x)=sinx(1+cosx)/sin^2x=(1+cosx)/sinx=右式