x-1分之(sin x)^2-(sin 1)^2的极限,x趋向于1

x-1分之(sin x)^2-(sin 1)^2的极限,x趋向于1 x*（1+sin^2 x )/sin^2x 不定积分 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 化简(sin x+cos x-1)(sin x-cos x+1)/sin 2x (sin 1/x) /x (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(sin^2 x/sin x-cosx)-(sin x+cosx/tan^2 x-1) ∫sin^2x/(1+sin^2x )dx求解, 求不定积分【sin(x)^5 + sin(x)^3】^(1/2) 1-2sinx*sin(60+x)sin(60-x) sin(x^2)/sin^2x 求导 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x 1/sin^2x的不定积分谢谢!（是1/sin x * sin x） 化简：2cos2x+2sin^2 x+cos(-x)分之sin2x+sin(π-x)=___________ sin(2x+1) 横坐标缩小到原来的2分之1 得到是什么?是sin(4x+2)还是sin(4x+1) 已知sin(x+4分之π)sin(4分之π-x)=6分之1,x∈(2分之π,π）,求sin4x的值 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