(2007•金山区一模)设函数f(x)=1+2cos(2x−π4)sin(x+π2).
设函数f(x)=cos(2x+π/3)+sin^2x-1/2
(2010•怀柔区一模)已知函数f(x)=1+2cos(2x−π4)sin(π2−x).
设函数f(x)=2sin(x−π4)cos(x+π4)+1,则f(x)是( )
(2010•长春三模)已知函数f(x)=sin(x+π2),g(x)=cos(x−π2),设h(x)=f(x)g(x),
设函数f(x)sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x) (1)求f(π/6),f(π/
已知函数f(x)=cos(2x-π/3)+sin^2x-cos^2x,设函数g(x)=[f(x)]^2+f(x),求g(
一,设函数f(x)=sin(πx/3-π/6)-2cos^2πx/6.
设函数f(x)=sin(πx/4-π/6)-2cos^2πx/8+1
设函数f(x)=sin(πx/4-π/6)-2cos²πx/8+1
设函数f(x)=cos(2x+π/3)+sin^2 X
设函数f(x)=cos(2x+π/3)+sin^2x
急.设函数f(x)=cos(2x+π/3)+sin^2 X