搜搜考试网求函数f(x)=√sinx ㏒2(9-x²) √-cosx的定义域
来源：学生作业帮助网 编辑：作业帮 时间：2024/07/07 04:34:35
f(x)=2sinx(sinX+cosX)=2sinxsinx+2sinxcosx=1-cos2x+sin2x=√2sin(2x-π/4)+1所以f(x)的最小正周期=2π/2=π最大值=1+√2
f(x)=|sinx|\2sinx+2cosx\|cosx|f(x=5/2(2kπ
先化简,f（x）=sin（2x-π/6）（1）最小正周期T=2π/2=π（2）x∈[0,π/4]2x-π/6∈[-π/6,π/3]∴f（x）单调递增最小值为f（0）=sin（-π/6）=-1/2最小值
再问：为什么sinx不为零
f(x)=2sinx(sinx+cosx)=2sin²x+2sinxcosx=1-cos(2x)+sin(2x)=√2sin(2x-π/4)+1当2kπ-π/2≤2x-π/4≤2kπ
f(x)=log2√(1+sinx)+log2√(1-sinx)=log2√(1+sinx)*√(1-sinx)=log2√[(1+sinx)*(1-sinx)]=log2√(cosx)^2=log2
我一会给你发过去f=2(sinx)^2+2sinxcosx=1-cos2x=sin2x=1+根号2(（根号2/2）*sin2x-(根号2/2)*cos2x)=1+根号2sin(2x-45°)所以最小正
f(x)=2sin^2x+2sinxcosx=sin2x-cos2x+1=sin(2x-π/4)+1因此最小正周期是π最大值是2
(1)已知向量a=(2cosx,sinx),b=(cosx,√3sinx)则f(x)=a*b=2cosx*cosx+sinx*√3sinx=2(cosx)^2+√3(sinx)^2=1+cos2x+√
f(x)=2sinxcosx+√3cos2x=sin2x+√3cos2x=2(sin2xcosπ/3+cos2xsinπ/3）=2sin(2x+π/3）f(x)最小正周期=2π/2=π
f(x)=2sinx(sinx+√3cosx)-1=2sin^2x+2√3sinxcosx-1=√3sin2x-cos2x=2sin(2x-π/6)1.函数的最小正周期=2π/2=π2.2kπ-π/2
是偶函数~因为,f(-x)=√(1-sin(-x))+√(1+sin(-x))=√（1+sinx)+√(1-sinx)=f(x),所以它是偶函数~
令t=sinx则f=√2(1-t^2)+5t-1=-√2(t^2-5t/√2)+√2-1=-√2(t-5√2/4)^2+33√2/8-1对称轴为t=5√2/4>1,开口向下因为|t|
f(x)=2sinx(sinx+cosx) =2sin²x+2sinxcosx =2sin²x-1+2sinxcosx+1&
f(x)=2sinx(sinx+cosx)=2(sinx)^2+2sinxcosx此时根据二倍角公式得=1+sin2x-cos2x用合一变形,其实你就是这一步不会,其实就是当：asinx+bcosx出
f(x)=2sinx(sinx+cosx)=2sin²x+2sinxcosx=1-cos2x+sin2x=sin2x-cos2x+1=√2sin(2x-π/4)+1当2x-π/4=2kπ+π
f(x)=sinx+sin(x+π/2)=sinx+cosx=√2sin(x+π/4)1.T=2π2.f(x)max=√2,f(x)min=-√23.f(a)=sina+cosa=3/4(sina)^
f(x)=1-2x^2