作业帮f(x)=(1+sinx+cosx)[sin(x/2)-cos(x/2)]/根号（2+2cosx）(1)f(x)的化简形式（最好有过程）
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/30 14:18:55
![f(x)=(1+sinx+cosx)[sin(x/2)-cos(x/2)]/根号（2+2cosx）(1)f(x)的化简形式（最好有过程）](/uploads/image/z/1043862-6-2.jpg?t=f%28x%29%3D%281%2Bsinx%2Bcosx%29%5Bsin%28x%2F2%29-cos%28x%2F2%29%5D%2F%E6%A0%B9%E5%8F%B7%EF%BC%882%2B2cosx%EF%BC%89%281%29f%28x%29%E7%9A%84%E5%8C%96%E7%AE%80%E5%BD%A2%E5%BC%8F%EF%BC%88%E6%9C%80%E5%A5%BD%E6%9C%89%E8%BF%87%E7%A8%8B%EF%BC%89)
f(x)=(1+sinx+cosx)[sin(x/2)-cos(x/2)]/根号（2+2cosx）(1)f(x)的化简形式（最好有过程）
f(x)=(1+sinx+cosx)[sin(x/2)-cos(x/2)]/根号（2+2cosx）
(1)f(x)的化简形式（最好有过程）
f(x)=(1+sinx+cosx)[sin(x/2)-cos(x/2)]/根号（2+2cosx）(1)f(x)的化简形式（最好有过程）
f(x)=(1+sinx+cosx)[sin(x/2)-cos(x/2)]/√（2+2cosx）
=[sin²(x/2)+cos²(x/2)+sinx+cosx][sin(x/2)-cos(x/2)]/[√2（1+cosx）]
=[sin²(x/2)+cos²(x/2)+2sin(x/2)*cos(x/2)+cos²(x/2)-sin²(x/2)][sin(x/2)-cos(x/2)]/[√4*（1+cosx）/2]
=｛[sin(x/2)+cos(x/2)]²+[cos(x/2)+sin(x/2)][cos(x/2)-sin(x/2)]｝[sin(x/2)-cos(x/2)]/[√4*cos²(x/2)]
=｛[sin(x/2)+cos(x/2)][sin(x/2)+cos(x/2）+cos(x/2)-sin(x/2)]｝[sin(x/2)-cos(x/2)]/[√4*cos²(x/2)]
=｛[sin(x/2)+cos(x/2)]*2cos(x/2）｝[sin(x/2)-cos(x/2)]/[2*cos(x/2)]
=｛[sin(x/2)+cos(x/2)][sin(x/2)-cos(x/2)]*2cos(x/2）｝/[2*cos(x/2)]
=[sin(x/2)+cos(x/2)][sin(x/2)-cos(x/2)]
=sin²(x/2)-cos²(x/2)
不明白的地方M我