已知函数f(x)=4sin(x+π/6)cosx-1,求f(π/12)
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/11/20 05:53:25
已知函数f(x)=4sin(x+π/6)cosx-1,求f(π/12)
需要FX的化简即可.
需要FX的化简即可.
f(x)=4sin(x+π/6)cosx-1
=4(sinxcosπ/6+cosxsinπ/6)cosx-1
=4(√3/2 sinx+1/2 cosx)cosx-1
=2√3sinxcosx+2cos^2 x-1
=√3*(2sinxcosx)+(2cos^2 x-1)
=√3sin2x+cos2x
=2(√3/2 sin2x+1/2cos2x)
=2(sin2xcosπ/6+cos2xsinπ/6)
=2sin(2x+π/6)
带入π/12
f(π/12) =2sin(2*π/12+π/6)=2sin(π/3)=2*√3/2=√3
和楼上结果一致,收工~
=4(sinxcosπ/6+cosxsinπ/6)cosx-1
=4(√3/2 sinx+1/2 cosx)cosx-1
=2√3sinxcosx+2cos^2 x-1
=√3*(2sinxcosx)+(2cos^2 x-1)
=√3sin2x+cos2x
=2(√3/2 sin2x+1/2cos2x)
=2(sin2xcosπ/6+cos2xsinπ/6)
=2sin(2x+π/6)
带入π/12
f(π/12) =2sin(2*π/12+π/6)=2sin(π/3)=2*√3/2=√3
和楼上结果一致,收工~
已知函数f(x)=sin(x+6分之π)+sin(x-6分之π)+cosx+a(a属于R,a是常数).(1)求函数f(x
已知函数f(x)=2sin(π -x)cosx,(1).求f(x)的最小正周期,(2.求f(x)在区间[-π/6,π/2
已知函数f(x)=4cos²x+sin²x-4cosx-2 (1)求f(π/3)的值.(2)求f(x
已知函数f(x)=2sin(π-x)cosx
已知函数f(x)=-根号(2)sin(2x+π/4)+6sinxcosx-2cosx^2+1,x属于R,求f(x)最小正
请帮忙解答一下 已知函数f(x)=2cosx*sin(x+π/3)-√3sin平方x+sinx*cosx 1 求函数f(
已知函数f(x)=sin(x+π/6)+sin(x- π /6)+cosx+a (x属于R,a为常数) ①求函数f(x)
已知函数f(x)=1-根号2sin(2x-π/4)/cosx
已知函数f(x)={1-(√2)*sin[2x-(π/4)]}/cosx
已知函数f(x)=2cos2x+sin^2x-4cosx.制值 (1)求f(π/3)的值 (2)求f
已知函数f(x)=sin(2x-π/6)+cosx,求sincos的值(若f(x)=1) 求单调增区间
已知函数f (x)=2sin(x+π/6)-2cosx,x∈【π/2,π】,求:函数f(x)的值域