已知函数f(x)=4cosxsin(x+派/6)-1,求f最小正周期,求其在区间[-派/6,派/4]上最大最小值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 20:57:57
![已知函数f(x)=4cosxsin(x+派/6)-1,求f最小正周期,求其在区间[-派/6,派/4]上最大最小值](/uploads/image/z/572278-22-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D4cosxsin%28x%2B%E6%B4%BE%2F6%29-1%2C%E6%B1%82f%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%2C%E6%B1%82%E5%85%B6%E5%9C%A8%E5%8C%BA%E9%97%B4%5B-%E6%B4%BE%2F6%2C%E6%B4%BE%2F4%5D%E4%B8%8A%E6%9C%80%E5%A4%A7%E6%9C%80%E5%B0%8F%E5%80%BC)
已知函数f(x)=4cosxsin(x+派/6)-1,求f最小正周期,求其在区间[-派/6,派/4]上最大最小值
已知函数f(x)=4cosxsin(x+派/6)-1,求f最小正周期,求其在区间[-派/6,派/4]上最大最小值
已知函数f(x)=4cosxsin(x+派/6)-1,求f最小正周期,求其在区间[-派/6,派/4]上最大最小值
由积化和差公式可得:
f(x)=2[sin(2x+π/6)+sin(π/6)]-1=2sin(2x+π/6)
故 f的最小正周期T=π
因为π/6
f(x)=4cosxsin(x+π/6)-1
=4cosx(sinxcosπ/6+cosxsinπ/6)-1
=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2sin(2x+π/6)
f(x)最小正周期T=2π/2=π
(2)
∵x∈[-π/6,π/4]
∴2x∈[-π/3...
全部展开
f(x)=4cosxsin(x+π/6)-1
=4cosx(sinxcosπ/6+cosxsinπ/6)-1
=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2sin(2x+π/6)
f(x)最小正周期T=2π/2=π
(2)
∵x∈[-π/6,π/4]
∴2x∈[-π/3,π/2]
∴2x+π/6∈[-π/6,2π/3]
∴2x+π/6=π/2时,f(x)max=2
2x+π/6=-π/6时,f(x)min=-1
收起
f(x)=4cosxsin(x+π/6)-1=4cosx(sinx*√3/2+cosx*1/2)-1
=2√3sinxcosx+2(cosx)^2-1
=√3sin2x+cos2x
=2*(sin2x*√3/2+cos2x*1/2)
=2sin(2x+π/6) ,
因此,最小正周期为 2π/2=π ,
当 -π/6<=x<=π/4 时,-π/6<=2...
全部展开
f(x)=4cosxsin(x+π/6)-1=4cosx(sinx*√3/2+cosx*1/2)-1
=2√3sinxcosx+2(cosx)^2-1
=√3sin2x+cos2x
=2*(sin2x*√3/2+cos2x*1/2)
=2sin(2x+π/6) ,
因此,最小正周期为 2π/2=π ,
当 -π/6<=x<=π/4 时,-π/6<=2x+π/6<=2π/3 ,
因此 -1/2<=sin(2x+π/6)<=1 ,
即函数在区间 [-π/6,π/4] 上的最大值为 2 ,最小值为 -1 。
收起