sinxcosx 根号下1 sinx的平方的积分

/>∵(1+tanx)/(1-tanx)=3+根号二∴1+tanx=3+√2-（3+√2)tanx∴(4+√2)tanx=2+√2∴tanx=(2+√2)/(4+√2)=(3+√2)/7∴cos

有sin(x-45°)=√2／4=sinxcos45°-cosxsin45°,得sinx-cosx=0.5,两边平方得1-2sinxcosx=0.25.sinxcosx=3／8.tanx+1／tanx

f(x)=cos^2x-sin^2x+2(根号3)sinxcosx+1=cos2x+(根号3)sin2x+1=2{(1/2)cos2x+[(根号3)/2]sin2x}+1=2sin(2x+派/6)+1

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

f(x)=5根号下3cos^2x+根号下3sin^2x-4sinxcosx=4根号下3cos^2x-4sinxcosx+根号下3=2根号下3(2cos^2x-1)-2sin2x+3根号下3=2根号3c

f(x)=5根号下3cos^2x+根号下3sin^2x-4sinxcosxx属于[π/4,7π/24]f(x)=根号3(5(1+cos2x)/2)+根号3(1-cos2x)/2-2sin2xf(x)=

sin^2X=1/2-(1/2)cos(2X))sof(x)=-1+cos(2x)+sqrt(3)sin2x+1=2sin(2x+pi/6)whenxbelongsto-pi/6,pi/3,2x+pi

f(x)=sin^2x+根号3*sinxcosx+1/2=（1-cos2x)/2+√3/2*sin2x+1/2=(√3/2*sin2x-1/2*cos2x)+1=sin(2x-π/3)+11)函数f(

(1+sina)/(1-sina)=(1+sina)²/(1-sina)(1+sina)=(1+sina)²/(1-sin²a)=(1+sina)²/cos&s

原式=√(1+sina)(1-sina)/(1-sina)²-√(1-sina)(1+sina)/(1+sina)²=｜cosa｜/(1-sina)-｜cosa｜/(1+sina)

y=sin²x+√3sinxcosx-1=[1-cos(2x)]/2+(√3/2)sin(2x)-1=(√3/2)sin(2x)-(1/2)cos(2x)-1/2=sin(2x-π/6)-1

∵√3/2-√3sin²x+sinxcosx=(√3/2)(1-2sin²x)+(1/2)sin2x=sin(π/3)cos2x+cos(π/3)sin2x=sin(2x+π/3)

解原式=1/2*（1-cos2x）+(根号3)/2*sin2x+1/2=sin2x*cos(派/6)-cos2x*sin(派/6)=sin(2x-派/6)(1)最小正周期=2派/ω=2派/2=派(2)

f(x)=sin^4x-2根号3sinxcosx-cos^4x+1=(sin^2x-cos^2x)(cos^2x+sin^2x)-2根号3sinxcosx+1=-cos2x-根号3sin2x+1=-2

(1+tanx)/(1-tanx)=3+2根号21+tanx=(3+2根号2)-(3+2根号)tanx2(2+根号2)tanx=2(根号2+1）tanx=(根号2+1)/(2+根号2)=根号2/2(s

1、∵1-cos2x=2sin^2∴f(x)=2sinxsinx+2√3sinxcosx+1=1-cos2x+√3sin2x+1=2+2(√3/2*sin2x-1/2*cos2x)=2(cos(π/6

1、f(x)=2sinxsinx+2√3sinxcosx+1=1-cos2x+√3sin2x+1=2sin(2x-π/6)+2对称轴：2x-π/6=kπ+π/2,得x=kπ/2+π/3对称中心：2x-

已知函数f(x)=2sin²x+2根号3sinxcosx+1,求函数的单调递增区间函数在区间【0,п/2】上的最值解析：∵函数f(x)=2sin²x+2√3sinxcosx+1=-

∫sinxcosx/(1+sin^4x)dx=∫sinx/(1+sin^4x)d(sinx)=1/2*∫1/(1+(sin^2x)^2)d(sin^2x)=1/2*arctan(sin^2x)+C

f(x)=√3sin2x+(1-cos2x)-1=√3sin2x-cos2x=2sin(2x-π/6)最小正周期T=2π/2=π单调增区间：2kπ-π/2=