已知sin(α+π/2)=-√5/5,α∈(0,π)求[cos2(π/4+α/2)-cos2(π/4-α/2)]/[sin(π-α)+cos(3π+α)]的值已知sin（α+π/2）=-√5/5,α∈（0,π）.求[cos2(π/4+α/2)-cos2(π/4-α/2)]/[sin(π-α)+cos(3π+α)]的值

已知sin(α+π/2)=-√5/5,α∈(0,π)求[cos2(π/4+α/2)-cos2(π/4-α/2)]/[sin(π-α)+cos(3π+α)]的值已知sin（α+π/2）=-√5/5,α∈（0,π）.求[cos2(π/4+α/2)-cos2(π/4-α/2)]/[sin(π-α)+cos(3π+α)]的值
已知sin(α+π/2)=-√5/5,α∈(0,π)求[cos2(π/4+α/2)-cos2(π/4-α/2)]/[sin(π-α)+cos(3π+α)]的值
已知sin（α+π/2）=-√5/5,α∈（0,π）.求[cos2(π/4+α/2)-cos2(π/4-α/2)]/[sin(π-α)+cos(3π+α)]的值

已知sin(α+π/2)=-√5/5,α∈(0,π)求[cos2(π/4+α/2)-cos2(π/4-α/2)]/[sin(π-α)+cos(3π+α)]的值已知sin（α+π/2）=-√5/5,α∈（0,π）.求[cos2(π/4+α/2)-cos2(π/4-α/2)]/[sin(π-α)+cos(3π+α)]的值
sin（α+π/2）=cosa=-√5/5 所以α∈（π/2,π）.sina=2√5/5
[cos2(π/4+α/2)-cos2(π/4-α/2)]/[sin(π-α)+cos(3π+α)]
=[(1+cos(π/2+α))/2-(1+cos(π/2-α))/2]/[sina-cosa]
=-sina/(sina-cosa)
=(-2√5/5)/(2√5/5+√5/5)
=-2/3

∵sin（α+π/2）=-√5/5
∴cosa=-√5/5 ∵α∈（0，π）.∴α∈（π/2，π﹚
sina=2√5/5
[cos2(π/4+α/2)-cos2(π/4-α/2)]/[sin(π-α)+cos(3π+α)]
=﹙-sina－sina﹚/﹙sina＋cosa﹚
=﹙-4√5/5﹚/﹙√5/5﹚
=-4老天哪个才是正确答案啊！~你这个式...

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∵sin（α+π/2）=-√5/5
∴cosa=-√5/5 ∵α∈（0，π）.∴α∈（π/2，π﹚
sina=2√5/5
[cos2(π/4+α/2)-cos2(π/4-α/2)]/[sin(π-α)+cos(3π+α)]
=﹙-sina－sina﹚/﹙sina＋cosa﹚
=﹙-4√5/5﹚/﹙√5/5﹚
=-4

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