已知y=sin(wx a)是偶函数-搜答案-搜搜考试网

sin^2(x-y)+sin^2(y-z)+sin^2(z-x)=[1-cos2(x-y)+1-cos2(y-z)+1-cos2(z-x)]/2=3/2-[(cos2xcos2y+sin2xsin2y

振幅为2；周期为π；初相为π/3单增区间：kπ－5π／12≦x≦kπ＋π／12对称轴：x=﹙1／2﹚kπ+（1／12）π

（0,1）代入原式知sinφ=1/2因为|φ|

设sinθ-cosθ＝√2sin(x-π/4)＝t则：t属于[-1,√2]sinxcosx=(1-t^2)/2y=-t^2+t+1=-(t-1/2)^2+5/4最大值是：5/4(此时t=1/2)最小值

sin(x+y)=1所以x+y=2kπ+π/2所以2x+y=2(x+y)-y=4kπ+π-y所以tan(2x+y)=tan(4kπ+π-y)因为tan的周期是π所以tan(4kπ+π-y)=tan[4

已知x,y,z都是锐角,sin²x+sin²y+sin²z=1,求tanx*tany*tanz的最值证明：由原式得1-cos²x+1-cos²y+1-

sina=3/2sin^2α+sin^2βy=sin^2α+sin^2β＝sin^2α＋sina－3/2sin^2α＝sina－1/2sin^2α-1=

∵sin（x+y）=1，∴x+y=π2+2kπ，k∈Z，∴y=-x+π2+2kπ，∴sin（2y+x）=sin（-2x+π+4kπ+x）=sin（π-x）=-sinx=-13，故答案为：-13．

y=sin²x+sinx+cosx+2=(1-cos2x)/2+√2sin(x+л/4)+2=(1/2)*sin(2x+л/2)+√2*sin(x+л/4)+5/2；=(1/2)*sin(2

我列个去,就算我高中毕业到现在已经8年了,我也看的出来1楼的乱说的撒,值域明显是[-2,2]嘛

x,y,z属于(0,派/2)sinx,cosx∈（0,1）对于a>0,b>0,有不等式：开根号下（a^2+b^2)≥根号2*(a+b)/2sin^2x+sin^2y+sin^2z=1cosx=开根号下

要求tanx/tany则x,y≠π/2+kπ,且y≠kπ(k为整数）,则cosxcosy≠0sin(x+y)/cos(x-y)=m/n(sinxcosy+cosxsiny)/(cosxcosy+sin

偶函数则x=0是对称轴sin的对称轴是在函数取最值得地方所以sin(0*w+q)=sinq=1或-10

首先得T/2=2π-3π/4=5π/4所以：T=5π/2,即2π/w=5π/2,所以：w=4/5；所以：y=sin(4x/5+A),把点(3π/4,-1)代入,得：-1=sin(-3π/5+A)所以：

因为,-π/2

y=cos2x+sin²x-cosx=cos²x-cosx=(cosx-1/2)²-1/4x=2kπ+π,max(y)=2x=2kπ±π/3,min(y)=-1/4x∈[

y=3sin(x+20°)+5sin(x+80°）=3sin(x+20°)+5sin(x+20°+60°）=3sin(x+20°)+5*[sin(x+20°)*cos60°+cos(x+20°)*si

方程y=sin(x+y)两边对x求导数有：y'=cos(x+y)(x+y)'=cos(x+y)(1+y')移项整理得：[1-cos(x+y)]y'=cos(x+y)因此：y'=cos(x+y)/[1-

任何正弦函数,只要系数是1,值域就是[-1,1]