已知3π/2
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已知3π/2已知3π/2已知3π/2√(1+sinx)-√(1-sinx)+√(2-2cosx)=|sin(x/2)+cos(x/2)|+|sin(x/2)-cos(x/2)|+√[2-2(1-4si
已知3π/2
已知3π/2
已知3π/2
√(1+sinx)-√(1-sinx)+√(2-2cosx)=|sin(x/2)+cos(x/2)|+ |sin(x/2)-cos(x/2)|+ √[2-2(1-4sin^2(x/2)]
=sin(x/2)+cos(x/2)+ sin(x/2)-cos(x/2)+ 2sin(x/2)
=4sin(x/2)
解
原式=|sin(x/2)+cos(x/2)|-|sin(x/2)-cos(x/2)|+2|sin(x/2)|
∵3π/2＜x＜2π
3π/4＜x/2＜π
∴sin(x/2)＞0＞cos(x/2)
∴原式
=sin(x/2)+cos(x/2)-sin(x/2)+cos(x/2)+2sin(x/2)
=2sin(x/2)+2cos(x/2)
=(2√2)sin[(x/2)+(π/4)]