已知sin(α+2/π)=-√5/5,α∈(0,π),求cos∧2(π/4+α/2)-cos∧2(π/4-α/2)/sin(π-α)+cos(3π+α)的值
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![已知sin(α+2/π)=-√5/5,α∈(0,π),求cos∧2(π/4+α/2)-cos∧2(π/4-α/2)/sin(π-α)+cos(3π+α)的值](/uploads/image/z/10177976-56-6.jpg?t=%E5%B7%B2%E7%9F%A5sin%28%CE%B1%EF%BC%8B2%2F%CF%80%29%EF%BC%9D-%E2%88%9A5%2F5%2C%CE%B1%E2%88%88%280%2C%CF%80%29%2C%E6%B1%82cos%E2%88%A72%28%CF%80%2F4%EF%BC%8B%CE%B1%2F2%29%EF%BC%8Dcos%E2%88%A72%28%CF%80%2F4-%CE%B1%2F2%29%2Fsin%28%CF%80%EF%BC%8D%CE%B1%29%EF%BC%8Bcos%283%CF%80%EF%BC%8B%CE%B1%29%E7%9A%84%E5%80%BC)
已知sin(α+2/π)=-√5/5,α∈(0,π),求cos∧2(π/4+α/2)-cos∧2(π/4-α/2)/sin(π-α)+cos(3π+α)的值
已知sin(α+2/π)=-√5/5,α∈(0,π),求cos∧2(π/4+α/2)-cos∧2(π/4-α/2)/sin(π-α)+cos(3π+α)的值
已知sin(α+2/π)=-√5/5,α∈(0,π),求cos∧2(π/4+α/2)-cos∧2(π/4-α/2)/sin(π-α)+cos(3π+α)的值
sin(α+2/π)=cosα=-√5/5;
∴α∈(π/2,π);
sinα=2√5/5;
cos^2(π/4+α/2);
=1/2[cos(π/2+α)+1];
=1/2(1-sinα);
cos^2(π/4-α/2);
=1/2[cos(π/2-α)+1];
=1/2(1+sinα);
原式
=[1/2(1-sinα)-1/2(1+sinα)]/(sinα-cosα);
=-sinα/(sinα-cosα);
=-2√5/5/(2√5/5+√5/5);
=-2/3;
因为sin(α+π/2)=-√5/5,α∈(0,π),所以cosα=-√5/5 sinα=2√5/5
cos∧2(π/4+α/2)-cos∧2(π/4-α/2)/sin(π-α)+cos(3π+α)=1/2(1-sin)-(1+sinα)/(2sinα)-cosα
=-(1/2)sinα-1/(2sinα)-cosα
=-(1/2)(2√5/5)-1/(4√5...
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因为sin(α+π/2)=-√5/5,α∈(0,π),所以cosα=-√5/5 sinα=2√5/5
cos∧2(π/4+α/2)-cos∧2(π/4-α/2)/sin(π-α)+cos(3π+α)=1/2(1-sin)-(1+sinα)/(2sinα)-cosα
=-(1/2)sinα-1/(2sinα)-cosα
=-(1/2)(2√5/5)-1/(4√5/5)-(-√5/5)
=-√5/4
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