1 sinx cosx在0-π 2积分

f（x）=cos^2x-2sinxcosx-sin^2x=-sin2x+cos2x=-√2*sin(2x-π/4)-π/2

y=√3sinxcosx+cos^2x-1/2=（√3/2）sin2x+（1/2）（cos2x+1）-1/2=（√3/2）sin2x+（1/2）cos2x=sin（2x+π/6）因为x属于[0,π/2

f(x)=2*3^(1/2)sinxcosx+2cos^2x+m=3^(1/2)sin2x+cos2x+m+1=2sin(2x+pai/6)+m+1xE[0,pai/2]时,请画图,f(x)max=2

f(x)=根号3sin2x+cos2x=2（cosπ/6sin2x+sinπ/6cos2x）=2sin（2x+π/6）因为函数在区间[0,π/2]上所以π/6≤2x+π/6≤7π/6当2x+π/6=π

f(x)=2倍根号下3sinxcosx+2cos^2(x-1﹚该式子错误!应为f(x)=2倍根号下3sinxcosx+2cos^2x-1＝√3sin2x＋cos2x＝2sin﹙2x＋π/6﹚,函数的最

1、f(x)=√3sinxcosx-cos^2x+1/2=√3/2*2sinxcosx-1/2*(2cos^2x-1)=√3/2*sin2x-1/2*cos2x=sin2x*cosπ/6-cos2x*

f(x)=-√3sin^2x+sinxcosx=-√3(1-cos2x)/2+sin2x/2=sin2x/2+√3cos2x/2-√3/2=sin(2x+π/3)-√3/2因为X∈[0,π/2],所以

y=(1/sinx)+(1/cosx)+(1/sinxcosx)=（sinx+cosx+1）/(sinxcosx)令t=sinx+cosx=根号2sin（x+π/4）∵x∈(0,π/2),∴t∈（1,

2*(sinxcosx-cos平方x)+1=2sinxcosx-2cos²x+1=2sinxcosx-cos2x=sin2x-cos2x=√2sin(2x-45°)

分析:将函数变成只含一个三角函数f(x)=-√3sin²x+sinxcosx=-√3×(1-cos(2x))/2+(sin(2x))/2(利用二倍角公式)=-√3/2+(√3/2)cos(2

第一步,f(x)化简得f(x)=2cos^2x/(sinxcosx-sin^2x)第二步,由定义域知cosx不为0,f(x)除以cosx的平方,得f(x)=2/(tanx-tan^2x)第三步,另g(

f(x)=sinxcosx-(sinx+cosx)=0.5[1-(sinx+cosx)]²-1sinx+cosx=根号2sin(x+π/4)函数f(x)在(0,π/2）上的单调递增区间是(π

y=2(sinx)^2-sinxcosx+3(cosx)^2=2-sin2x/2+(cosx)^2=2-sin2x/2+cos2x/2+1/2=5/2+(√2/2)cos(2x+π/4)π+2kπ

(1-2sinxcosx)(1+2sinxcosx)=(sin²x+cos²x-2sinxcosx)(sin²x+cos²x+2sinxcosx)=(sinx-

f(x)=√3sinxcosx-(sinx)^2-3/2=(√3/2)*sin2x-(1-cos2x)/2-3/2=(√3/2)*sin2x+(1/2)*cos2x-2=sin(2x+π/6)-2x∈

F[x]=sinxcosx+cos^2x-1/2=1/2sin2x+1/2(cos2x+1)-1/2=1/2(sin2x+cos2x)=√2/2sin(2x+π/4)最小正周期T=2π/W=π2x+π

答：f(x)=2√3sinxcosx+2(cosx)^2-1=√3sin(2x)+cos(2x)=2*[(√3/2)*sin(2x)+(1/2)cos(2x)]=2sin(2x+π/6)最小正周期T=

f(x)=√2sin(2x+π/4)=√2sin2(x+π/8),x在（-3π/8,π/8）是增函数,在（π/8,5π/8）时减函数；f(x)在[0,π/2]上的最大值为x=π/8时,此时f(x)=√

f(x)=1-cos2x+√3sin2x+1=2(√3/2sin2x-1/2cos2x)+2=2sin(2x-π/6)+2∵x∈[0,π/2]∴2x-π/6∈[-π/6,5π/6]∴sin(2x-π/