sin(1-x)/x^2-4x+3 lim趋向1

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 (1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x 化简[1-(sin^4 x-sin^2 xcos^2 x+cos^4 x)]/(sin^2 x)+3sin^2 x 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x s = 2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x)用matlab画图,求教啊 三角等式求证：cos^6x+sin^6x=1-3sin^2x+3sin^4x 求证(3-sin^4 x-cos^4 x)/2cos^2 x=1+tan^2 x+sin^2 x 化简2sin^2[(π/4)+x]+根号3（sin^x-cos^x）-1 求极限 （（sin(x^3+x^2-x)+sin x) /x x→0 已知lim sinx/x=1 求不定积分【sin(x)^5 + sin(x)^3】^(1/2) 急,matlab解非线性方程组用什么函数呢 方程如下所示：(cos(x(1))*cos(x(3))-sin(x(1))*cos(x(2))*sin(x(3)))*(194-x(4))*x(7)+(cos(x(1))*sin(x(3))+sin(x(1))*cos(x(2))*cos(x(3)))*(145-x(5))*x(7)+sin(x(1))*sin(x(2))*x(6)*x(7)-1=0;(-sin(x(1 泰勒公式的为什么㏑( 1 + sin X ) = sin X - ( sin X )²/2 +(sin X )³ /3 -(sin X )∧4 + o (sin ∧4 X ),完全不理解, 求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x 求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π/4-x/2). 已知sin(x+π/6)=1/4,求sin(5π/6-x)+sin^2(π/3-x) 已知 Sin(X+ 派/6)=1/4求 Sin(5/6派 - X)+Sin^2 (派/3 -X) x^ 3(sin x )^2/x^ 4+2x +1在[-1,1]的定积分 (sin 1/x) /x