题目

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）

提问时间：2020-10-05

答案

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