若270
问题描述：
若270
数学人气：361 ℃时间：2020-01-25 21:59:58
优质解答
=[2sinθ/2cosθ/2+2(cosθ/2)^2]*(sinθ/2-cosθ/2)/√[4(cosθ/2)^2]
=2cosθ/2*(sinθ/2+cosθ/2)(sinθ/2-cosθ/2)/2|cosθ/2|
=cosθ/2*(-cosθ)/|cosθ/2|
∵(270＜θ＜360,135＜θ/2＜180 )
∴=cosθ
=2cosθ/2*(sinθ/2+cosθ/2)(sinθ/2-cosθ/2)/2|cosθ/2|
=cosθ/2*(-cosθ)/|cosθ/2|
∵(270＜θ＜360,135＜θ/2＜180 )
∴=cosθ
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答
=[2sinθ/2cosθ/2+2(cosθ/2)^2]*(sinθ/2-cosθ/2)/√[4(cosθ/2)^2]
=2cosθ/2*(sinθ/2+cosθ/2)(sinθ/2-cosθ/2)/2|cosθ/2|
=cosθ/2*(-cosθ)/|cosθ/2|
∵(270＜θ＜360,135＜θ/2＜180 )
∴=cosθ