求y=3sin(2x π 6)的单调区间
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sin(2x+π/3)∈(-1,1)sin(2x+π/3)+2∈(1,3)∴函数y=log3的定义域为R值域为单调性:单调递增,单调递减周期性:π最值:最大值:1,最小值:0
设u=1-2sin(2X+π/3),U>0,这就是罪(2X+π/3)]再问:麻烦不要复制好么==
把两个三角函数展开,得y=3/2sinx-√3/2cosx合并成:y=√3sin(x-π/6)单调区间是(-π/3,2π/3)增(2π/3,5π/3)减其中都要加上2kπ,我就不写了
增区间-π/2+2kπ
∵x∈(-π/6,π); ∴2x+π/3∈(0,2π+π/3); 则函数y的最大值为1,最小值为-1; 则y∈【-1,1】
y=3sin(π/6-x/2)=-3sin(x/2-π/6)的单调递增区间是y=3sin(x/2-π/6)的减区间令2kπ+π/2≥x/2-π/6≥2kπ+3π/24kπ+2π/3≥x≥4kπ+5π/
y=|sin^22x|=|(1-2sin^22x)/2-1/2|=|(cos^22x-sin^22x)/2-0.5|=|0.5cos4x-0.5|最小正周期是pie/2|sin^22(-x)|=|0.
y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos
sin^2x+cos^2y=1/2∴sin^2x=1/2-cos^2y3sin^2x+sin^2y=3(1/2-cos^2y)+sin^2y=1.5-3cos^2y)+sin^2y又有sin^2y+c
y=sin(x+π/3)sin(x+π/2)=sin(x+π/3)cosx=(sinxcosπ/3+cosxsinπ/3)cosx=1/2sinxcosx+√3/2cos^2(x)[cos^2(x)指
先把y=sinx的图像的纵坐标不变,横坐标缩小为原来的1/3,得到y=sin3x的图像,再将y=sin3x的图像的横坐标保持不变,纵坐标扩大为原来的2倍,得到y=2sin3x的图像,最后,将y=2si
[-k¥-¥5/8,-k¥-¥/8]
y=sin(2x+π/3)+cos(2x-π/6)=(1/2)sin2x+(√3/2)cos2x+(√3/2)cos2x+(1/2)sin2x=sin2x+√3cos2x=2sin(2x+π/3)2k
利用相关法因为sinx在[2kpi-pi/2,2kpi+pi/2]上递增,在[2kpi+pi/2,2kpi+3pi/2]上递减所以让(2x+pi/2)属于[2kpi-pi/2,2kpi+pi/2],也
周期为2pai/2=pai最值为3单调性为(-1/3pai,1/6pai)递增(1/6pai,2/3pai)递减
y=sin(-3x)=-sin3x单调递增区间是2kPai+Pai/2
y=sinx增区间[2kπ-π/2,2kπ+π/2]所以本题,2kπ-π/2≤π/4+2x≤2kπ+π/2kπ-3π/8
任何正弦函数,只要系数是1,值域就是[-1,1]