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
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![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](/uploads/image/z/13302952-16-2.jpg?t=%E5%B7%B2%E7%9F%A5sin%28%CE%B1%2B%CF%80%2F2%29%3D-%E2%88%9A5%2F5%2C%CE%B1%E2%88%88%280%2C%CF%80%29%E6%B1%82%5Bcos2%28%CF%80%2F4%2B%CE%B1%2F2%29-cos2%28%CF%80%2F4-%CE%B1%2F2%29%5D%2F%5Bsin%28%CF%80-%CE%B1%29%2Bcos%283%CF%80%2B%CE%B1%29%5D%E7%9A%84%E5%80%BC%E5%B7%B2%E7%9F%A5sin%EF%BC%88%CE%B1%2B%CF%80%2F2%EF%BC%89%3D-%E2%88%9A5%2F5%2C%CE%B1%E2%88%88%EF%BC%880%2C%CF%80%EF%BC%89.%E6%B1%82%5Bcos2%28%CF%80%2F4%2B%CE%B1%2F2%29-cos2%28%CF%80%2F4-%CE%B1%2F2%29%5D%2F%5Bsin%28%CF%80-%CE%B1%29%2Bcos%283%CF%80%2B%CE%B1%29%5D%E7%9A%84%E5%80%BC)
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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