f(x)=(x+sinx)/cosx,(-π/2
1个回答
是奇函数;
f(-x)=(-x+sin(-x))/cos(-x) =(-x -sinx)/cosx = -(x+sinx)/cosx = -f(x)
相关问题
f(x)=4sin^2[(π+2x)/4].sinx+(cosx+sinx)(cosx-sinx)
f(x)=2sinx(sinx+cosx),x∈[-π/2,π/2]时,求f(x)的值域
求解:f(x)=2sinx*(π/2-x)-√3sin(π+x)cosx+sin(π/2+x)cosx
化简f(x)=4sinx*sin^2((π+2x)/4)+(cosx+sinx)(cosx-sinx)
设f(x)满足f(-sinx)+3f(sinx)=4sinx•cosx(|x|≤[π/2]).
设f(x)满足f(-sinx)+3f(sinx)=4sinx•cosx(|x|≤[π/2]).
判断f(x)=(1+sinx-cosx)/(1+cosx+sinx)(-π/2<x<π/2)的奇偶性
已知函数f(x)=cos(2x-π/3)+(cosx+sinx)(cosx-sinx)
f(x)=sinx+sin(x+π/2)=sinx+cosx怎样变的?
f(x)=cosx√((1-sinx)/(1+sinx))+sinx√((1-cosx)/(1+cosx))求f(π/4