y=-2tan(3x pai 4)的单调递减
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令a=tanx则a属于Ry=f(x)=(a-a+1)/(a+a+1)ya+ya+y=a-a+1(y-1)a+(y+1)a+(y-1)=0a是实数则方程有解所以判别式大于等于0(y+1)-4(y-1)>
sin(x+y)=sinxcosy+cosxsiny=1/2sin(x-y)=sinxconsy-cosxsiny=1/3sinxcosy=5/12,cosxsiny=1/12tanx/tany=si
tan,正切;sin,正弦;cos,余弦tan(x+y)tan(x-y)=sin(x+y)/cos(x+y)*sin(x-y)/cos(x-y)=sin(x+y)sin(x-y)/[cos(x+y)c
tanx函数的周期是π,所以y=tan(2x-3)的周期等于π除以2=π/2
∵y=tan(2x-π3),∴其周期T=π2.
一.(tanβ-tanα)/(1-tanβ*tanα)=-2∵tanα=1/3∴(tanβ-1/3)/(1-tanβ*1/3)=-2tanβ-1/3=-2(1-1/3*tanβ)3tanβ-1=-6+
你看后面TAN里一个x一个x+y那你就把给你的原式中的2x+y拆开,在消消化化的,试下吧我觉得能行
y'=1/(tan(x/2))*(tan(x/2))'=1/(tan(x/2))*(sec^2(x/2))*(x/2)'=1/(2sin(x/2)*cos(x/2))=1/sin(x)=csc(x)
令a=x+y,则条件变为3sin(a-x)=sin(a+x),展开得3sinacosx-3cosasinx=sinacosx+cosasinx,移项2sinacosx=4cosasinxtana=2t
y=3tan(π6-x4)=-3tan(x4-π6),∴T=π|ω|=4π,∴y=3tan(π6-x4)的周期为4π.由kπ-π2<x4-π6<kπ+π2,得4kπ-4π3<x<4kπ+8π3(k∈Z
已知sin(x+y)=1,求证:tan(2x+3y)=tany证明:sin(x+y)=1所以x+y=2k兀+兀/2K为整数所以tan(2x+3y)=tan(4k兀+兀+y)=tan(兀+y)=tany
1,函数y=tan(x+兀/3)的对称中心为x+兀/3=k兀x=k兀-兀/3对称中心为(k兀-兀/3,0)k∈Z2,求函数y=-2tan(3x+兀/3)的定义域,值域,并指出它的周期,奇偶性和单调性定
周期为pai/2定义域为集合2X-pai/4不等于kpai+pai/2k属于整数单调递增区间为kpai-pai/2
sin[(x+y)+x]=5sin[(x+y)-x]sin(x+y)·cosx+cos(x+y)·sinx=5·sin(x+y)·cosx-5·cos(x+y)·sinx4·sin(x+y)·cosx
sin(x+2y)=3sinx,sin[(x+y)+y]=3sin[(x+y)-y],sin(x+y)cos(y)+cos(x+y)sin(y)=3[sin(x+y)cos(y)-cos(x+y)si
函数y=tan^2(x)-2tan(x),=(tanx-1)^2+1-60°
设一个变量u=y/x,带入方程很好求解,解不出来再联系我哈
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(1)由sin(2α+β)=3sinβ,得sin[(α+β)+α]=3sin[(α+β)-α],sin(α+β)cosα+cos(α+β)sinα=3sin(α+β)cosα-3cos(α+β)sin
cos(2x+y)=3cosycos(x+y+x)=3cos(x+y-x)cos(x+y)cosx-sin(x+y)sinx=3[cos(x+y)cosx+sin(x+y)sinx]2cos(x+y)