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

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

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

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)=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

(1) f(x)=(1+cos2x)/2 -(1/2)sin2x -1/2=(1/2)(cos2x-sin2x)=(√2/2)[(√2/2)cos2x-(√2/2)sin2x]=(√2/2)sin(π/4 -2x)从而,最小正周期为T=2π/2=π,值域为[-√2/2,√2/2](2)f(a/2)=(√2/2)sin(π/4-a)=3√2/10所以

f(x)=cos2x+sin2x=√2sin(2x+π/4)则最小正周期为2π/2=π再令2x+π/4=π/2+2kπ 得x=π/8+kπ (k为任意常数)即x=π/8+kπ 时f(x)有最大值,最大值为√2

解f(x)=cosx-sinx+2sinxcosx =sin2x+cos2x =√2(√2/2sin2x+√2/2cos2x) =√2(sin2xcosπ/4+cos2xsinπ/4) =√2sin(2x+π/4)∴最小正周期为:T=2π/2=πx∈[-π/4,π/4]∴2x+π/4∈[-π/4,3π/4]∴当2x+π/4=π/2时取得最大值为:√2∴x=π/8

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)=2sinxcosx+2cosx=2sinxcosx+(2cosx-1)+1,=sin2x+cos2x+1=(√2)sin(x+(π/4))+1

(1)f(x)=√3sinxcosx+cosx=√3/2sin2x+(cos2x+1)/2=√3/2sin2x+1/2cos2x+1/2=sin(2x+π/6)+1/2f(x)的最小正周期为2π/2=π2kπ-π/2≤2x+π/6≤2kπ+π/2 (k∈Z)kπ-π/3≤x≤kπ+π/6则f(x)的单调递增区间为[

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!

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