求函数最小正周期{要有过程}
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求函数最小正周期{要有过程}
FX=sin{2x+PI/6}+sin{2x-pi/6}+cos2x
FX=sin{2x+PI/6}+sin{2x-pi/6}+cos2x
![求函数最小正周期{要有过程}](/uploads/image/z/16780550-14-0.jpg?t=%E6%B1%82%E5%87%BD%E6%95%B0%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%7B%E8%A6%81%E6%9C%89%E8%BF%87%E7%A8%8B%7D)
f(x)=[sin(2x+π/6)+sin(2x-π/6)]+cos(2x)
=2sin{[(2x+π/6)+(2x-π/6)]/2}cos{[(2x+π/6)-(2x-π/6)]/2}+cos(2x)
=2sin(2x)cos(π/6)+2cos(2x)sin(π/6)
=2sin(2x+π/6)
式中振幅A=2,角频率ω=2,初相φ0=π/6,又sinx的最小正周期T0=2π,所以
f(x)的最小正周期T=T0/ω=(2π)/2=π
=2sin{[(2x+π/6)+(2x-π/6)]/2}cos{[(2x+π/6)-(2x-π/6)]/2}+cos(2x)
=2sin(2x)cos(π/6)+2cos(2x)sin(π/6)
=2sin(2x+π/6)
式中振幅A=2,角频率ω=2,初相φ0=π/6,又sinx的最小正周期T0=2π,所以
f(x)的最小正周期T=T0/ω=(2π)/2=π