化简【(cos2/x)/根号(1+sinx)】+【(sinx/2)/根号(1-sinx)】(x∈(3π/2,2π)
问题描述：
化简【(cos2/x)/根号(1+sinx)】+【(sinx/2)/根号(1-sinx)】(x∈(3π/2,2π)
答
x∈(3π/2,2π)
x/2∈(3π/4,π)
(cosx/2)/√(1+sinx)+(sinx/2)/√(1-sinx)
=[(cosx/2)√(1-sinx)+(sinx/2)√(1+sinx)]/√[(1-sinx)(1+sinx)].通分
=[cosx/2*|sinx/2-cosx/2|+sinx/2*|sinx/2+cosx/2|]/|cosx|
=[cosx/2*(sinx/2-cosx/2)-sinx/2*(sinx/2+cosx/2)]/(cosx)
=[-cos²x/2-sin²x/2]/cosx
=-1/cosx