(cos(x/2)/(根号1+sinx))+(sin(x/2)/根号(1-cosx)),x属于(3/2π,2π)

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(cos(x/2)/(根号1+sinx))+(sin(x/2)/根号(1-cosx)),x属于(3/2π,2π)
(cos(x/2)/(根号1+sinx))+(sin(x/2)/根号(1-cosx)),x属于(3/2π,2π)

(cos(x/2)/(根号1+sinx))+(sin(x/2)/根号(1-cosx)),x属于(3/2π,2π)
∵sin(x/2)=±√[(1-cosx)/2]
cos(x/2)=±√[(1+cosx)/2
又∵x∈(3π/2,2π)
∴3π/4


1+sinx=(sinx/2+cosx/2)^2
1-cosx=2【sin(x/2)】^2
原式=cos(x/2)/(sinx/2+cosx/2)+sin(x/2)/√2sin(x/2)
=1/(tanx/2+1)+√2/2
x属于(3/2π,2π)
原式值域在(√2/2,1+√2/2)
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