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
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/25 22:02:23
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...
全部展开
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
收起