根号下1-2sin(π+2)cos(π+2)等于多少?

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根号下1-2sin(π+2)cos(π+2)等于多少?
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根号下1-2sin(π+2)cos(π+2)等于多少?
根号下1-2sin(π+2)cos(π+2)等于多少?

根号下1-2sin(π+2)cos(π+2)等于多少?
√[1-2sin(π+2)cos(π+2)]
=√[sin^2(π+2)+cos^2(π+2)-2sin(π+2)cos(π+2)]
=√[sin(π+2)-cos(π+2)]^2
=|sin(π+2)-cos(π+2)|
=|cos2-sin2|
=sin2-cos2
≈sin114.64° -cos114.64 °
=1.326

=√1-2sin2cos2=√(sin2-cos2)^2=︱sin2-cos2︱=sin2-cos2(2rad是第二象限角)

√(1-2sin(π+2)cos(π+2))=√((sin(π+2))^2+(cos(π+2))^2)-2sin(π+2)cos(π+2))=√((sin(π+2)-cos(π+2))^2)=sin(π+2)-cos(π+2)=√2sin(5π/4+2)