2sin^2(2x)+4sin2xcos2x+3cos^2(2x)最小正周期和单调区间?计时开始...honey..

为何sin2xcosπ/6-cos2xsinπ/6=2sin(2x-π/6) 已知函数f(x)=1/2sin2xcosφ+sin²xsinφ+…… =2*(sin2xcosπ/6+cos2xsinπ/6) 怎么等于 =2sin(2x+π/6) sin(2x+α）=sin(-2x+α）的展开整理详细过程,我知道结果是sin2Xcosα=0 已知y=sin(cosx)^2*cos(sinx)^2,求y'答案是-sin2xcos(cos2x), f(x)=sin2x—cos2x①=sin2xcos(π/4)—cos2xsin(π/4)②=√2sin(2x-π/4)③这里面的第二步是怎么来的,求解 函数Y=SIN2X+COS2X的值域?解：y=sin2x+cos2x=√2(sin2x*√2/2+cos2x*√2/2) =√2[sin2xcos(π/4)+cos2xsin(π/4)] =√2sin（2x+π/4) 因为 sinx∈[-1,1] (为什么 sinx∈[-1,1）谢谢sin（2x+π/4)∈[-1,1] 所以 √2sin（2x+π/4) √3sin2x+cosx为什么等于2（sin2xcosπ/6+cos2xsinπ/6）其中的√3去哪了?cosπ/6和sinπ/6是怎么来的? 老师这题求详解!我数学底子差 看不懂是怎么变式的f(x)=√3sinxcosx-cos²x-1/2=√3/2(2sinxcosx)-1/2(1+cos2x)-1/2=√3/2sin2x-1/2cos2x-1=sin2xcosπ/6-cos2xsinπ/6-1=sin(2x-π/6)-1故 f(x)的最小正周期是 π,最小值是 -2. 已知函数f(x)=sin2xcosφ-2cos²xsin(π-φ)-cos(π/2+φ) 1.化简y=f(x)的表达式并求函数的周期2.当-π/2 已知函数f(x)=sin2xcosφ-2cos²xsin(π-φ)-cos(π/2+φ) （-π/2＜φ＜π/2）已知函数f(x)=sin2xcosφ-2cos²xsin(π-φ)-cos(π/2+φ) x=派/6时取最大值1.求φ的值2.将函数Y=f(x)图像上各点的横坐标扩大到原来2倍, 化简sin^4x-sin^2x+cos^2x 化简:sin^4 x-sin^2 x+cos^2 x 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 2sin(x+π/4)sin(x-π/4)=? sin(x^2)/sin^2x 求导