函数f(x)=sin(2x+π/6)cos(2x-π/3)的最小正周期为?
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函数f(x)=sin(2x+π/6)cos(2x-π/3)的最小正周期为?函数f(x)=sin(2x+π/6)cos(2x-π/3)的最小正周期为?函数f(x)=sin(2x+π/6)cos(2x-π
函数f(x)=sin(2x+π/6)cos(2x-π/3)的最小正周期为?
函数f(x)=sin(2x+π/6)cos(2x-π/3)的最小正周期为?
函数f(x)=sin(2x+π/6)cos(2x-π/3)的最小正周期为?
利用公式cosα=sin(π/2+α)
cos(2x-π/3)=sin(2x-π//3+π/2)=sin(2x+π/6)代入:
f(x)=sin²(2x+π/6)=1/2-(1/2)cos(4x+π/3);
T=2π/ω=2π/4=π/2
利用积化和差公式:sin(2x+pi/6)cos(2x-pi/3)=[sin(4x-pi/6)+sin(pi/2)]/2
所以最小正周期为2pi/4=pi/2
T=2π/ω=2π/4=π/2
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