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已知函数F(x)=sinxCosx+Cos²x 求函数F(x)的最小正周期,并写出...

f(x)=sinxcosx+cosx=(1/2)(sin2x+1+cos2x)=(√2/2)sin(2x+π/4)+1/2,周期是π,对称轴方程是2x+π/4=(k+1/2)π,k∈Z.f(x)=2x-3(m+1)x+6mx,m∈R,f'(x)=6x^2-6(m+1)x+6m=6(x-1)(x-m),m=1时f'(x)>=0,f(x)是增函数;m≠1时x介于1,m之间时f'(x)<0,f(x)是减函数;其余的情况,f(x)是增函数.

f(x)=2sinx(sinx+cosx)=2(sinx)^2+2sinxcosx=1-cos2x+sin2x=√2sin(2x-π/4)+1所以函数fx的最小正周期是T=2π/2=π如果不懂,请追问,祝学习愉快!

f(x)=sinxcosx+cos方x=1/2sin2x+(1+cos2x)/2=1/2 (sin2x+cos2x)+1/2=√2/2 sin(2x+π/4)+1/2所以(1)求函数f(x)的最小正周期 最小正周期=π(2)求f(x)的最小值及相应的x的取值最小值=-√2/2 +1/2此时2x+π/4=2kπ+3π/22x=2kπ+5π/4x=kπ+5π/8

F(x)=(cosx+sinx)(cosx-sinx)=cosx-sinx =cos2x f(x)的最小正周期T=2π/ω=2π/2=π

有些符号打不出来,见谅了!f(x)=sinxcosx+cos^2=1/2sin2x+1/2(cos2x+1)=1/2(sin2x+cos2x)+1/2=根号2/2sin(2x+π/4)+1/2所以周期为π;当x∈【0,π/2】时;2x+π/4∈【π/4,3π/4】则当2x+π/4=π/2时,f(x)有最大值,为

f(x)=(sinx+cosx)+2cosx=sinx+2sinxcosx+cosx+2cosx=1+2sinxcosx+2cosx=2sincosx+2cosx-1+2=sin2x+cos2x+2.正弦,余弦二倍角公式=√2sin(2x+π/4)+2.辅助角公式令π/2+2k

此题需要用到二倍角公式和辅助角公式sinxcosx=1/2sin2x√3cosx=√3cosx+√3/2-√3/2=√3(cosx-1/2)+√3/2=√3/2cos2x+√3/2∴f(x)=1/2sin2x+√3/2cos2x+√3/2=sin(2x+π/3)+√3/2∴ T=2π/2=π

已知f(x)=sinxcosx+cosx∴f(x)=1/2sin2x+(cos2x+1)/2=1/2sin2x+1/2cos2x+1/2=√2/2*(√2/2*sin2x+√2/2*cos2x)+1/2=√2/2*[sin(2x+45°)]+1/2∴T=2π/ω=2π/2=π∴f(x)的最小正周期为π

f(x)=sinx-cosx =√2sin(x-π/4)最小正周期为 2π

f(x)=sinxcosx+cosx=sin2x/2+cos2x/2+1/2=1/根号(2)*sin(2x+pi/4)+1/2T=2pi/w=pif(x)MAX=根号(2)/2+1/2 此时 2x+pi/4=pi/2+2k*pi ∴x=pi/8+k*pif(x)MIN=-根号(2)/2+1/2 此时 2x+pi/4=-pi/2+2k*pi ∴x= -3pi/8+k*pi当 -pi/2+2k*pi=

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