求心脏线r=1-sin(f)的面积

f(x)=cos(-x/2)+sin(π-x/2)=cosx/2+sinx/2=√2(√2/2sinx/2+√2/2cosx/2)=√2sin(x/2+π/4)最小正周期T=2π/(1/2)=4π增区

(1)f(x)=sin(x+7π/4)+cos(x-3π/4)=sin(x+7π/4)+sin(5π/4-x)=2sin(3π/2)cos(x+π/4)=-cos(x+π/4)最小正周期T=2π/1=

由于函数关系式已知,（1）第一问可直接用T=2π/ω得到.ω=2,则最小正周期T=π（2）因为最小正周期为π,且图像向右平移π/4个单位长度.所以在(π/4,5π/4)上有完整函数图像.可得最大值2在

心脏线关于x轴（极轴）对称,只需一半的曲线即可,即可令0≤θ≤π；V=∫π(ρsinθ)²dx={0,2π/3}∫π(ρsinθ)²d(ρcosθ)-{2π/3,π}∫π(ρsin

【参考答案】r＝1+cosθ,r'＝-sinθ利用对称性长度=2∫(0,π)√r^2+r'^2dθ=2∫(0,π)√(2+2cosθ)dθ=2∫(0,π)√4cos^2(θ/2)dθ=4∫(0,π)c

f(x)=2sin^2(x+π/4)-√3cos2x-1=-cos(2x+π/2)-√3cos2x=sin2x-√3cos2x=2sin(2x-π/3)当x属于R时,函数f(x)的最小正周期T=2π/

由题得sin(2x-π/4)=-1/2所以2x-π/4=2kπ-5π/6或2x-π/4=2kπ-π/6解得x=kπ-7π/24或x=kπ+π/24（k∈N）

再答：请给好评，不懂可追问

1.f(x)=(sinx)^2+2sinxcosx+3(cosx)^2=1+sin2x+2(cosx)^2=sin2x+cos2x+2=√2sin(2x+π/4)+2所以f(x)的最大值是2+√2当2

f(x)=[2sin(x+π/3)+sinx]cosx-√3sin^2x=[sinx+√3cosx+sinx]cosx-√3sin^2x=2sinxcosx+√3cos^2x-√3sin^2x=sin

(1)f(x)=cos(-x/2)+sin(π-x/2)=cos(x/2)+sin(x/2)=√2[(√2/2)cos(x/2)+(√2/2)sin(x/2)]=√2[sin(π/4)cos(x/2)

因为f(x)=根号3sin(2x-π/6)+2sin的平方（x-π/12)=根号3sin(2x-π/6)-（1-2sin的平方（x-π/12)）+1=根号3sin(2x-π/6)-cos(2x-π/6

f(x)=sin(2x+π/6)+2msinxcosx=(√3sin2x)/2+(cos2x)/2+msin2x=(m+√3/2)sin2x+(cos2x)/2所以f(x)的最大值为√[1/4+(m+

f(x)=1－cos²x＋cosx+5/8-3/2,设t=cosx,t∈[－1,1]这样,y=1－t²＋t＋5/8－3/2,这是个二次函数的区间最值问题.

f(x）=sinxsin(xπ/2)=sinxcosx故f(x)的最小正周期是2πf^2(x）=(sinxcosx)^2=1sin2xsin2x=-7/16sin(xπ/2)=cosxf

f(x)=2cos^2x*sin^2x=1/2(sin2x)^2=(1-cos4x)/4T=2π/w=π/2f(-x)=(1-cos(-4x))/4=f(x)即f(x)为偶函数f(x)的最小周期为π/

试试看：如图所示：

等等,一会给你,我也算出和答案不一样,不知怎么回事,照片是过程,再问：我也是这个答案哎！再答：可能是答案有问题吧，做法又没有错，采纳吧啊啊

1、f(x)=cosx-(-sinx)=sinx+cosx=√2(√2/2*sinx+√2/2cosx)=√2(sinxcosπ/4+cosxsinπ/4)=√2sin(x+π/4)所以最大值=√22

解f(x)=2sin^2x-sin(π+2x),=2sin^2x+sin(2x)=2（1-cos2x）/2+sin(2x)=1-cos2x+sin(2x)=sin(2x)-cos2x+1=√2sin（