已知f(x)=[sin(-nπ-x)cos(nπ+x)]/cos[(n+1)π-x]×1/tan(x-nπ) (n∈Z),求f(7π/6)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/18 23:43:04
![已知f(x)=[sin(-nπ-x)cos(nπ+x)]/cos[(n+1)π-x]×1/tan(x-nπ) (n∈Z),求f(7π/6)](/uploads/image/z/3812341-13-1.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%3D%5Bsin%28-n%CF%80-x%29cos%28n%CF%80%2Bx%29%EF%BC%BD%2Fcos%5B%28n%2B1%29%CF%80-x%EF%BC%BD%C3%971%2Ftan%28x-n%CF%80%29+%28n%E2%88%88Z%29%2C%E6%B1%82f%287%CF%80%2F6%29)
x){}K4*4m34t7Vh&k Y5
5Ar@
K4*@54utDi<ؔa~AL&Hff&.xwrQtB lu6U
QkMhB5@}1XjBچ(&ۂ1
M2J@! H
已知f(x)=[sin(-nπ-x)cos(nπ+x)]/cos[(n+1)π-x]×1/tan(x-nπ) (n∈Z),求f(7π/6)
已知f(x)=[sin(-nπ-x)cos(nπ+x)]/cos[(n+1)π-x]×1/tan(x-nπ) (n∈Z),求f(7π/6)
已知f(x)=[sin(-nπ-x)cos(nπ+x)]/cos[(n+1)π-x]×1/tan(x-nπ) (n∈Z),求f(7π/6)
①当n=2k时,
f(x) =-sinxcosx/(-cosx)·(1/tanx)
=-cosx
f(7π/6)=√3/2
②当n=2k+1时,
f(x)=sinx·(-cosx)/cosx·(1/tanx)
=-cosx
f(7π/6)=√3/2
∴当n∈Z时,f(7π/6)=√3/2
已知函数f(x)=sinπx/3(x∈N),f(1)+f(2)+.+f(99)=( )
已知函数f(x)=sinπx/2 , 则∑1-2007【f(n)+f'(n)】=?
已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n∈Z) 求f(π/2010)+f(502π/1005)已知f(x)=cos²(nπ+x)sin²(nπ-x)/cos²[(2n+1)π-x](n∈Z)求f(π/2010)+f(502π/1005)
已知f(x)=sin
已知f(x)=sin(nπ-x)cos(nπ+ x)/cos[(n+1)π-x]*tan(x-nπ)(n属于Z)化简f(x)
已知函数F(X)=SIN(2X+φ)(-π
已知函数f(x)=cos2x/[sin(π/4-x)]
π已知f(x)=sin(nπ-x)cos(nπ+ x)/cos[(n+1)π-x]*tan(x-nπ)(n属于Z)求f(6/7π)
已知f(x)=[sin(-nπ-x)cos(nπ+x)]/cos[(n+1)π-x]×1/tan(x-nπ) (n∈Z),求f(7π/6)
已知函数f(x)=2根号3sin平方x-sin(2x-π/3)
已知函数f(x)=(1+1 anx)sin^2x+m sin(x+π/4)sin(x-π/4)
对于任意实数x和整数n,已知f(sinx)=sin(4n+1)x,求f(cosx)f(cosx)=f(sin(90°-x))=sin(4n+1)(90°-x)=sin[360°n+90°-(4n+1)x] =sin[90°-(4n+1)x]=cos(4n+1)x 以下这几步看不懂sin[360°n+90°-(4n+1)x] =sin[90°-(4n+1)x]
已知x∈R ,n∈Z,且f(sinx)=sin(4n+1)x,则f(cosx)=
已知x∈R,n∈Z,且f(sinx)=sin(4n+1)x,则f(cosx)=?
对任意实数x和整数n,已知f(sinx)=sin(4n+1)x,求f(cosx)
已知x属于实数,n属于整数,且f(sinx)=sin(4n+1)x,求f(cosx).
已知F(X)=根号3COS^2 X+SIN XCOS X-2SIN X*SIN(X-π/6),求F(X)的最大值
f(x)=sin(nπ/3),f(1)+f(2)+...+f(2010)等于