高一三角函数化简题(1+sinθ+cosθ)×(sin(θ/2)-cos(θ/2))——————————————————
高一三角函数化简题
(1+sinθ+cosθ)×(sin(θ/2)-cos(θ/2))
——————————————————
根号下（2+2cosθ）
由于等级不够,不能上图所以大家体谅下哈.中间是分数线
(1+sinθ+cosθ)×(sin(θ/2)-cos(θ/2))
——————————————————
根号下（2+2cosθ）
由于等级不够,不能上图所以大家体谅下哈.中间是分数线
赞 共回答了15个问题采纳率：80% 举报
1+sinθ+cosθ
=sin²(θ/2)+cos²(θ/2)+2sin(θ/2)cos(θ/2)+cos²(θ/2)-sin(θ/2)
=[sin(θ/2)+cos(θ/2)]²+[sin(θ/2)+cos(θ/2)][cos(θ/2)-sin(θ/2)]
=[sin(θ/2)+cos(θ/2)][sin(θ/2)+cos(θ/2)+cos(θ/2)-sin(θ/2)]
=2cos(θ/2)[sin(θ/2)+cos(θ/2)]
2+2cosθ
=2+2[2cos²(θ/2)-1]
=4cos²(θ/2)
所以分母=2|cos(θ/2)|
所以原式=2cos(θ/2)[sin(θ/2)+cos(θ/2)][sin(θ/2)-cos(θ/2)]/2|cos(θ/2)|
=-cos(θ/2)[cos²(θ/2)-sin²(θ/2)]/|cos(θ/2)|
=-cos(θ/2)cosθ/|cos(θ/2)|
=cosθ或-cosθ
=sin²(θ/2)+cos²(θ/2)+2sin(θ/2)cos(θ/2)+cos²(θ/2)-sin(θ/2)
=[sin(θ/2)+cos(θ/2)]²+[sin(θ/2)+cos(θ/2)][cos(θ/2)-sin(θ/2)]
=[sin(θ/2)+cos(θ/2)][sin(θ/2)+cos(θ/2)+cos(θ/2)-sin(θ/2)]
=2cos(θ/2)[sin(θ/2)+cos(θ/2)]
2+2cosθ
=2+2[2cos²(θ/2)-1]
=4cos²(θ/2)
所以分母=2|cos(θ/2)|
所以原式=2cos(θ/2)[sin(θ/2)+cos(θ/2)][sin(θ/2)-cos(θ/2)]/2|cos(θ/2)|
=-cos(θ/2)[cos²(θ/2)-sin²(θ/2)]/|cos(θ/2)|
=-cos(θ/2)cosθ/|cos(θ/2)|
=cosθ或-cosθ
1年前
9
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