求y=sin(π 6-4x)的单调递增区间

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

y=sinxcos30+cosxsin30-cosxsin60-sinxcos60=sinx[(根号3-1)/2]+cosx[(1-根号3)/2]=[(根号3-1)/2](sinx-cosx)=[(根

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==

y=sin²x+sin2x+2cos²x=sin2x+(1+cos2x)/2+1=sin2x+1/2*cos2x+3/2=√5/2(2/√5*sin2x+1/√5*cos2x)+3

y=sin^4x+cos^4x=sin^4x+cos^4x+2sin^2xcos^2x-2sin^2xcos^2x=（sin^2x+cos^2x）^2-2sin^2xcos^2x=1-1/2sin^2

∵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=sinx在（2kπ-π/2,2kπ+π/2)为增,在(2kπ+π/2,2kπ+3π/2)为减函数y=cosx在(2kπ-π,2kπ）为增,在（2kπ,2kπ+π）为减函数y=tanx在（kπ-π/

（1）横坐标伸长到原来的3倍则函数变为y=sin(2x+π/4)(x系数变为原来的1/3）（2）向右平移π/8个单位则函数变为y=sin[2(x-π/8)+π/4]=sin2x

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'sin(y/x)-y/x*sin(y/x)+1=0令y/x=u,则y'=u+xu'所以(u+xu')sinu-usinu+1=0xu'sinu+1=0-sinudu=dx/x两边积分：cosu=l

（1）因为x∈(-π/2,π/2),则x+π/4∈(-π/4,3π/4)所以由正弦函数的单调性可知：函数y=sin(x+π/4),x∈(-π/2,π/2)的值域是（-√2/2,1]（2）函数y=1/2

[-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

y=sin(x+π/6)sin(x-π/6)+acosx=(3/4)(sinx)^2-(1/4)(cosx)^2+acosx=-(cosx)^2+acosx+3/4=-(cosx-a/2)^2+a^2

y=sin(x+π/6)sin(x-π/6)+acosx=-1/2[cos(x+π/6+x-π/6)-cos(x+π/6-x+π/6)+acosx=-1/2(cos2x-cosπ/3)+acosx=-

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]