设k∈Z函数y=sinπ 4 x 2
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![设k∈Z函数y=sinπ 4 x 2](/uploads/image/f/7249302-54-2.jpg?t=%E8%AE%BEk%E2%88%88Z%E5%87%BD%E6%95%B0y%3Dsin%CF%80+4+x+2)
你写错了吧应该是f(x)=sin(x-θ)+√3cos(x-θ)则f(x)=√(1+3)*sin(x-θ+z)其中tanz=√3/1=√3z=π/3所以f(x)=2sin(x-θ+π/3)sin对称轴
记a=x+y,b=xtan(x+y)=2tanxsina/cosa=2sinb/cosb2sinbcosa=sinacosb3siny-sin(2x+y)=3sin(a-b)-sin(a+b)=2si
(1)因为x,y∈Z,又因为k∈Z,不妨设x=k,y=k-1有x^2-y^2=k^2-(k-1)^2=2k-1因为M={a│a=x^2-y^2,x,y∈Z}所以2k-1∈M(k∈Z)(2)M={a│a
ZU(0,2π)f(z)=0.5/π[0,2π]f(z)=0其它zf(z)为Z的概率密度函数.Z的期望E(Z)=π,Z的方差D(Z)=π^2/3.E(X)=∫(0,2π)sinzf(z)dz=0.5/
由题意可知f(x1)=f(x)min=-1=>sin(π/2x1+π/3)=-1=>π/2x1+π/3=2k1π-π/2=>x1=1/(4k1-5/3)同理f(x2)=f(x)max=1=>sin(π
∂z/∂x=cos(x-y)∂z/∂y=-cos(x-y)dz=∂z/∂x*dx+∂z/∂y*dy=co
先对x求偏导数得z'(x)cosz=yz+z'(x)y所以z'(x)=yz/(cosz-y)同理对y求偏导数得z'(y)=xz/(cosz-x)所以dz=yz/(cosz-y)dx+xz/(cosz-
f(x)=sin2(x+y/2)由于sin2x对称轴为π/4+kπ/2;故x+y/2=π/4+kπ/2x=π/4+kπ/2-y/2;将x=x=π/8代入,得y=π/4+kπ,根据y的范围可知:y=-3
p=sin(kπ/3-π)=-sin(π-kπ/3)=-sinkπ/3
dz=Z'xdx+Z'ydy=2xcos(x^2+y^2)dx+2ycos(x^2+y^2)dy
再问:啊不好意思搞错了。。是z=e^(x^2+y^2),求dz,谢谢你帮我解答一下吧。。再答:
公式输入了好半天,希望可以看懂哈!另外,可以不用辅助函数,直接利用已知等式计算求导.
对称轴k*π/6*1/5+π/3=π/2+nπ,n为整数k=30n-5任意整数区间出现一个最大最小值,说明函数周期要小于等于12π/(k/5)=10π所以k最小取值为55
令-π/2+2kπ≤2x+φ≤π/2+2kπ,得:-π/4-φ/2+kπ≤x≤π/4-φ/2+kπ那么有-π/4-φ/2+kπ=-5π/12+kπ,π/4-φ/2+kπ=π/12+kπ,解得:φ=π/
由题当x=π/6时,f(x)=±3即(k/5)*(π/6)+π/3=2mπ±π/2即k=60m+5或k=60m-25(m∈z)又由最大值与最小值之间距离最少为T/2
y=sin(π/4+x/2)sin(π/4-x/2)=sin(π/4+x/2)sin[π/2-(π/4+x/2)]=sin(π/4+x/2)cos(π/4+x/2)=1/2sin(π/2+x)=1/2
(1)f(x)=cos(-12x)+cos(4k+12π−12x)=cos12x+cos(2kπ+12π−12x)=sin12x+cos12x=2sin(12x+π4),所以,f(x)的最小正周期T=
z=sin(x²y²)+3x-5y²+1所以δz/δx=cos(x²y²)*2xy²+3δz/δy=cos(x²y²)*
y=sin(π/4+x/2)sin(π/4-x/2)=-1/2(cosπ/2-cosx)=1/2(cosx)递增区间是x∈[2kπ+π,2(k+1)π]k∈Z