y=2^k-tan(x^2 x) inπ,求dy dx,dy.
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![y=2^k-tan(x^2 x) inπ,求dy dx,dy.](/uploads/image/f/908701-61-1.jpg?t=y%3D2%5Ek-tan%28x%5E2+x%29+in%CF%80%2C%E6%B1%82dy+dx%2Cdy.)
周期T=π/2由kπ-π/2<2x-π/3<kπ+π/2,k∈z得kπ/2-π/12<X<kπ/2+5π/12,k∈z单调增区间为(kπ/2-π/12,kπ/2+5π/12)
(1)由f(x)=tanx是奇函数-tanx=tan(-x)y=-tan(x+π/6)+2=tan(-x-π/6)+2令z=-x-π/6y=tanz+2的定义域为(-π/2+2kπ,π/2+2kπ)k
tan周期是π这里x系数是2所以T=π/2x=5π/12+kπ/2是不是求值?y=-tan(2x+π/3)=-tan(5π/6+kπ+π/3)=-tan(5π/6+π/3)=-(tan5π/6+tan
y=tan^2x+tanx+1=(tanx+1/2)^2+3/4>=3/4x∈R,且x不等于180k+90y∈[3/4,正无穷)
设tanx=ty=t^2+t+1=(t+1/2)^2+3/4x属于Rtanx从正无穷到负无穷都可取值所以转化为二次函数求值域下面不用我说了吧答案是【3/4,正无穷)
tanx周期是π这里x系数是2所以周期=π/2
记a=x+y,b=xtan(x+y)=2tanxsina/cosa=2sinb/cosb2sinbcosa=sinacosb3siny-sin(2x+y)=3sin(a-b)-sin(a+b)=2si
一般情况下,ω是正值.如果它是负值,可以先把他变成正数就是了.其实,ω的正负其实不影响他的周期. 例如:y=sin(-|ω|x+θ)+K=-sin(|ω|x-θ)+K,它的周期还是T=2π/ω下面哪
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
y=tanx周期为π.Y=tan(wx+β)周期为π/w这题.周期=π/2
y=tan^2x+tanx+1=(tanx+1/2)^2+3/4x∈R且x≠kπ+π/2时tanx∈R所以y=(tanx+1/2)^2+3/4>=3/4即y∈[3/4,+∞)
显然有tanx=tan(x+kπ),此为三角函数的基本常识.因为标准三角函数y=tanx的最小正周期为π,其定义域为x≠π/2+kπ/2(k∈z)所以对于函数y=tan(2x-(π/3))的应该满足2
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)
(1)k^2-1=0,且k+1=0,k+7≠0时为一元一次方程,解得:k=正负1(2)k^2-1=0时,但k+1≠0,k+7≠0时二元一次方程,解得:k=1
1.y=x/tanx当x->0时,极限是1,x=0是第一类可去间断点;当x->kπ(k≠0)时,极限是∞,x=kπ(k≠0)第二类可去间断点;当x->kπ+π/2时,极限是0,x=kπ+π/2是第一类
kπ-π/2<x+π/6<π/2+kπkπ-2π/3<x<kπ+2π/3所以定义域为(kπ-2π/3,kπ+2π/3)不懂问我
y=tanx周期是π,所以y=tan(2x-π/3)的周期是π/2.和正弦,余弦求周期类似.
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
1.y=(tan^2x-csc^2x)/(tan^2x+cot^2x-1)=(sin^2x/cos^2x-1/sin^2x)/(sin^2x/cos^2x+cos^2x/sin^2x-1)=(sin^