化简根号下1+cos(3π-θ)/2 (3π/2<θ<2π)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/31 22:16:36
![化简根号下1+cos(3π-θ)/2 (3π/2<θ<2π)](/uploads/image/z/1344503-47-3.jpg?t=%E5%8C%96%E7%AE%80%E6%A0%B9%E5%8F%B7%E4%B8%8B1%2Bcos%283%CF%80-%CE%B8%29%2F2+%283%CF%80%2F2%EF%BC%9C%CE%B8%EF%BC%9C2%CF%80%29)
x){3gv>dGvr~s;4@L}{$7h$铡KΆ\۞uL|{֎>_7e{˹j+o4KHP$"PcA9& e`m4\L.V\Z`d
OvBw#ĉ
Ov/IYMH!,t;H>x1lH$ف ^v
化简根号下1+cos(3π-θ)/2 (3π/2<θ<2π)
化简根号下1+cos(3π-θ)/2 (3π/2<θ<2π)
化简根号下1+cos(3π-θ)/2 (3π/2<θ<2π)
我们先计算根号里面的
1+cos(3π/2-θ/2)=1-cos(π/2-θ/2)=1-sin(θ/2)=1-2sin(θ/4)cos(θ/4)=[sin(θ/4)-cos(θ/4)]²
因为3π/2<θ<2π
所以3π/8<θ/4<π/2
所以sin(θ/4)>cos(θ/4)
所以1+cos(3π/2-θ/2)=√[sin(θ/4)-cos(θ/4)]²=sin(θ/4)-cos(θ/4)