-2sin(-x+m-π/3)=-2sin(x+m-π/3) sin(-x+m-π/3)=sin(x+m-π/3) 所以-x+m-π/3=2kπ+x+m-π/3或-x+m-π/以上过程是为什么

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 已知函数f(x)=(1+1 anx)sin^2x+m sin(x+π/4)sin(x-π/4) 为啥sin(-x+m-π/3)=sin(x+m-π/3) -x+m-π/3=2kπ+x+m-π/3,-x+m-π/3=2kπ+π-(x+m-π/3) -2sin(-x+m-π/3)=-2sin(x+m-π/3) sin(-x+m-π/3)=sin(x+m-π/3) 所以-x+m-π/3=2kπ+x+m-π/3或-x+m-π/以上过程是为什么 设f(x)=4cos(ωx-π/6)sinωx-cos(2ωx+π),其中ω>0,求函数y=f(x)的值域,请看问题补充f（x）=4cos（ωx-π/6）sinωx-cos（2ωx+π） =4(coswxcosπ/6+sinwxsinπ6)sinwx+cos2wx =2√3sinwxcoswx+2sin²wx+cos2wx =√3si 若sin(π+x)+cos(π/2+x)=-m,则cos（3π/2-π）+2sin（6π-x）=? 已知sin X+cos X=m,|m|小于等于根号2且|m|不等于1,求sin X^3+cos X^3,sin X^4+cos X^4 .若函数f(x)sin(2x+m)-π f(x)=1+sin(2x)-根号3 cos2x,|f(x)-m| f(x)=1+sin(2x)-根号3 cos2x |f(x)-m| 方程sin(x+π/3)=m/2,在[0,π]上有两个解,求实数m的范围 已知α为锐角,且sin²α-sinαcosα-2cos²α=0 (1)求tanα的值 (2)求si已知α为锐角,且sin²α-sinαcosα-2cos²α=0(1)求tanα的值(2)求sin(α-π/3)的值求详解 带拉格朗日余项的麦克劳林公式的sin和cos展开项的问题sin(x)的麦克劳林展开式sin(x) = x - x^3 / + x^5 / + ...+ ((-1)^(m-1))*((x^(2m-1)) / (2m - 1)!) + ((-1)^(m))*(cos(θx)*(x^(2m+1)) / (2m + 1)!)cos(x) = 1 - x^2 / + x^4 / + . 若sin x=（1-m）/（2m+3）,且x∈[-π/6,π/6],则m的取值范围是 已知(sinx)2-(siny)2=m求sin(x y)sin(x-y)求sin(x+y)sin(x-y) 若tanx=2,则2sin平方x-3sin（π-x）sin（π/2-x）= 已知sin(x+π/6)=1/3,求sin(5π/6-x)+sin^2(π/3-x) sin^2(π/3-x)+sin^2(π/6+x)