高一三角函数化简题(1+sinθ+cosθ)×(sin(θ/2)-cos(θ/2))——————————————————
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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
![高一三角函数化简题(1+sinθ+cosθ)×(sin(θ/2)-cos(θ/2))——————————————————](/uploads/image/z/6512281-25-1.jpg?t=%E9%AB%98%E4%B8%80%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E5%8C%96%E7%AE%80%E9%A2%98%281%2Bsin%CE%B8%2Bcos%CE%B8%29%C3%97%28sin%28%CE%B8%2F2%29-cos%28%CE%B8%2F2%29%29%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94)
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
高一三角函数化简题(1+sinθ+cosθ)×(sin(θ/2)-cos(θ/2))——————————————————
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