考拉宝
幼苗
共回答了20个问题采纳率:85% 举报
f(x)=cos(2x-π/3)-cos2x-1
=cos2x*cos(π/3)+ sin2x*sin(π/3) - 2cos2x*cos(π/3)-1
= - [cos2x*cos(π/3)- sin2x*sin(π/3)] -1
=-cos(2x+π/3)-1
1年前
追问
3
xjw198
举报
=cos2x*cos(π/3)+ sin2x*sin(π/3) - 2cos2x*cos(π/3)-1 这部前面看的懂,后面2cos2x*cos(π/3)怎么来的啊
举报
考拉宝
因为cos(π/3)=1/2,那么:2cos(π/3)=1,所以:cos2x=2cos2x*cos(π/3) 我这样处理是为了下一步化简的便利哈。
xjw198
举报
= - [cos2x*cos(π/3)- sin2x*sin(π/3)] -1 那这步又是怎么跳到的