√(1-2sin(π+2)cos(π+2))等于
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√(1-2sin(π+2)cos(π+2))等于
√(1-2sin(π+2)cos(π+2))等于
√(1-2sin(π+2)cos(π+2))等于
原式=√[1-2sin2cos2]
=√(sin²2+cos²2-2sin2cos2)
=√(sin2-cos2)²
=|sin2-cos2|
π/2<2<π
第二象限
sin2>0>cos2
所以原式=sin2-cos2
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√(1-2sin(π+2)cos(π+2))等于
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