设函数f(x)=cos(2x-π/3)-cos2x-1
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设函数f(x)=cos(2x-π/3)-cos2x-1
怎么化简?
怎么化简?
![设函数f(x)=cos(2x-π/3)-cos2x-1](/uploads/image/z/6120699-51-9.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dcos%282x-%CF%80%2F3%EF%BC%89-cos2x-1)
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
再问: =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
再答: 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。
再问: = - [cos2x*cos(π/3)- sin2x*sin(π/3)] -1 那这步又是怎么跳到的
=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
再问: =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
再答: 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。
再问: = - [cos2x*cos(π/3)- sin2x*sin(π/3)] -1 那这步又是怎么跳到的
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