函数f(x)＝根号cos(sinx

(1)f(x)=sinxcosx+√3cos²X-√3/2=sin2x/2+√3cos2x/2+√3/2-√3/2=sin(2x+π/3).(2)f(x)的最小正周期为π,值域是[-1,1]

1,f(x)=sin²x+√3sinxcosx+2cos²x＝1－cos²x＋√3/2sin2x+2cos²x＝cos²x＋√3/2sin2x+1=(

已知:函数f(x)=2sinxcosx+2√3cos²x-√3求:(1)单调增区间和最小正周期；(2)当x∈[-π/4,π/4]时求最值.f(x)=2sinxcosx+2√3cos²

f(x)=2sinxcosx+2√3cos²x-√3=2sinxcosx+√3(2cos²x-1)=sin2x+√3cos2x=2sin(2x+π/3)最小正周期T=2π/2=π,

1f(x)=√3sinπx+cosπx=2（（√3/2）sinπx+（1/2）cosπx）=2sin（πx+π/3）∴最小正周期T=2π/w=2π/π=2值域f（x）∈[-2,2]2-π/2+2kπ＜

解f(x)=√3cos²x+sinxcosx-√3/2=√3*(1+cos2x)/2+(1/2)sin2x-√3/2=(1/2)sin2x+(√3/2)cos2x=sin(2x+π/3)∴T

(1)f(x)=2(cosx)^2+√3sin2x=cos2x+√3sin2x+1=2sin(2x+π/6)+1maxf(x)=3minf(x)=-1(2)f(x)=1-√32sin(2x+π/6)+

f(x)＝根号3×sin(x/4)cos(x/4)＋(cosx/4)＾2＝根号3／2＊sinx/2+(cosx/2+1)/2＝sinx/2cosPai/6+sinPai/6cosx/2+1/2=sin

求导得：f′（x）=-4sinxcosx+23cos2x=-2sin2x+23cos2x=4sin（π3-2x），令f′（x）=0，得到x=π6，∵f（0）=2+a，f（π2）=a，f（π6）=3+a

题目呢,不全,只有已知

y=sinx^2+根3sinxcosx+2cosx^2=-1/2(1-2sinx^2)+1/2根3*2sinxcosx+2cosx^2-1+3/2=-1/2cos2x+二分之根3倍sin2x+cos2

（Ⅰ）f（x）=m•n＝cos2ωx−sin2ωx+23cosωx•sinωx=cos2ωx+3sin2ωx=2sin(2ωx+π6)∵ω＞0∴函数f（x）的周期T=2π2ω＝πω，由题意可知T2≥π

f(x)=sin2x-2√3(cosx)^2+√3=sin2x-√3(1+cos2x)+√3=sin2x-√3cos2x=2sin(2x-π/3)π/4=再问：π/6=

f(x)=sinxcosx+√3(cosx)^2-√3/2=(1/2)sin2x+(√3/2)cos2x=sin2xcosπ/3+cos2xsinπ/3=sin(2x+π/3)1.0

只要cos(sinx)≥0就行了而sinx∈[-1,1]即-1弧度到1弧度在这个范围内余弦值始终为正所以定义域是R

偶函数

f(x)=sinxcosx+√3(cosx)^2-√3/2=(1/2)sin2x+(√3/2)cos2x=sin2xcosπ/3+cos2xsinπ/3=sin(2x+π/3)

f（x）=√3cos²x+sinxcosx-√3/2=√3(cos2x+1)/2+sin2x/2-√3/2=√3/2cos2x+√3/2+1/2sin2x-√3/2=1/2sin2x+√3/

f(x)=cosx+sinx=√2(√2/2*sinx+√2/2cosx)=√2(sinxcosπ/4+cosxsinπ/4)=√2sin(x+π/4)所以：f(x)的最大值=√2f(a)=cosa+

f(x)={[(sinx)^2+(cosx)^2]^2-(sinxcosx)^2}/(2-sin2x)=[1-(sinxcosx)^2]/(2-2sinxcosx)=(1+sinxcosx)(1-si