1-2sin^2(x+π/4)+2sin(x+π/8)cos(x+π/8)
=1-sin^2(x+π/4)-sin^2(x+π/4)+sin（2x+π/4)
=cos^2(x+π/4)-sin^2(x+π/4)+sin（2x+π/4)
=cos（2x+π/2)+sin（2x+π/4)
=sin（2x+π/4)-sin2x
所以最小正周期T=2π/2=π
fx=1-2sin^2(x+π/4)+2sin(x+π/8)cos(x+π/8)
2个回答
相关问题
-
f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)=cos(2x+π/4)+sin(
-
已知函数f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)
-
已知函数f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)
-
已知函数fx=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
-
1-2sin（x+8分之π）+2sin（x+8分之π）cos（x+8分之π）
-
已知函数f（x）=1-2sin平方（x+π/8）+2sin（x+π/8）点乘cos（x+π/8）
-
设函数f（x）=sin（[πx/4−π6]）-2cos2[πx/8]+1．
-
设函数f(x)＝sin(πx4−π6)−2cos2πx8+1
-
设函数f(X)=sin(πx/4-π/6)-2cos^2 π/8x+1
-
设函数f(x)=sin(πx/4-π/6)-2cos^2πx/8+1