作业帮f(x)=(6sin^4x-7sin^2x+2)/(sin^2x-cos^2x)
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/03 17:23:47
f(x)=(6sin^4x-7sin^2x+2)/(sin^2x-cos^2x)
f(x)=(6sin^4x-7sin^2x+2)/(sin^2x-cos^2x)
f(x)=(6sin^4x-7sin^2x+2)/(sin^2x-cos^2x)
化简：f(x)=(6sin⁴x-7sin²x+2)/(sin²x-cos²x)
f(x)=(2sin²x-1)(3sin²x-2)/(sin²x-cos²x)=[sin²x-(1-sin²x)](3sin²x-2)/(sin²x-cos²x)
=(sin²x-cos²x)(3sin²x-2)/(sin²x-cos²x)=3sin²x-2
(-1 - 3 Cos2x)/2
f(x)=(6sin^4x-7sin^2x+2)/(sin^2x-cos^2x)
=(2sin^2x-1)(3sin^2-2)/[sin^2x-(1-sin^2x)]
=3sin^2x-2
=3[(2sin^2x-1)+1]/2
=3(-cos2x+1)/2
=3(1-cos2x)/2
f(x)=(6sin^4x-7sin^2x+2)/(sin^2x-cos^2x)
傅里叶级数作图f（x）=2sin[x] - sin[2x] + 2/3sin[3x] - 1/2sin[4x]我用mathematica输入程序Plot[{2sin[x],-2sin[x],2sin[x] - sin[2x],-2sin[x] + sin[2x],2sin[x] - sin[2x] + 2/3sin[3x],-2sin[x] + sin[2x] - 2/3sin[3x],2sin[x] - sin[2x] + 2/3si
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