tan(π 4-x)怎么展开
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 08:46:04
1.左=tan(x/2+π/4)+tan(x/2-π/4)=tan[(x/2+π/4)+(x/2-π/4)][1-tan(x/2-π/4)tan(x/2+π/4)]=tanx[1-(-1)]=2tan
equ=sym('tan(x)=4*x/(x^2+4)');x=solve(equ);>>xx=0再问:这只能求出一个解啊再答:还有其他解吗?
lim(x->0)tan(x+πsinx/(4x))=tan(0+π/4)=1
(tanx)^(-4)*secx=(cosx)^3/(sinx)^4∴∫(tanx)^(-4)*secxdx=∫[1-(sinx)^2]d(sinx)/(sinx)^4=-(cscx)^3/3+csc
x=tan(π/4)=1,即x=1.这是一条过点(1,0)的、平行于y轴的直线.其倾斜角为90度.
∫(tan²x+tan⁴x)dx=∫tan²x(1+tan²x)dx=∫tan²xsec²xdx=∫tan²xdtanx=(1/
根据高一的公式y=tan(π/2-x)=cotx值域是[-1,0)U(0,1]准对!
tanx定义域是(kπ-π/2,kπ+π/2)则kπ-π/2
tanx的递增区间是(-π/2+kπ,π/2+kπ)-π/2+kπ
tan(X/2+π/4)+tan(x/2-π/4)=(tanx/2+1)/(1-tanx/2)+(tanx/2-1)/(1+tanx/2)=[(tanx/2+1)^2-(tanx/2-1)^2]/[(
∵tanx的单调增区间为(2kπ-π2,2kπ+π2)∴函数f(x)=tan(x+π4)的单调增区间为2kπ-π2<x+π4<2kπ+π2,即kπ−3π4<x<kπ+π4(k∈Z)故答案为(kπ−3π
∵tan(x+87π)=tan(x+π+π7)=tan(x+π7)=t,∴sin(157π+x)+3cos(x−137π)sin(207π−x)−cos(x+227π)=sin(x+π7+2π)+3c
1)π/4
证明:左边=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]
设tan(x+π/4)=t则t属于(-∞,+∞)当t=2值域是(-∞,-2]并[2,+∞)因为y=t+1/t在(-∞,-1)并(1,+∞)上是单调递增的而tan(-π/4+kπ)=-1tan(π/4+
分子把平方展开之后整个式子化为4tan(x/2)/[1-(tan(x/2))^2]=2{tan(x/2)+tan(x/2)/[1-(tan(x/2))×(tan(x/2))]}=2tanx再问:。。=
tan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)ta
secx+tanx=1/cosx+sinx/cosx=(1+sinx)/cosxtan(π/4+x/2)=[tanπ/4+tan(x/2)]/[1-tan(x/2)]=[1+tan(x/2)]/[1-
lim(x→0)[tan(π/4-x)]^(cotx)=lim(x→0){e^[cotx*ln(tan(π/4-x))]}只需要求lim(x→0)[cotx*ln(tan(π/4-x))];lim(x