已知sin(α+2/π)=-√5/5,α∈(0,π),求cos∧2(π/4+α/2)-cos∧2(π/4-α/2)/si
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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
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;
∴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;
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