已知tan(x+π/4)=3 ,则sinxcosx=

sinxcosx=1/2·sin2x=-1/2·cos(2x+π/2)
设x+π/4=p
所以，tanp=3
cos(2x+π/2)
=cos2p
=（1-tan²p）/（1+tan²p） 万能公式
=(1-9)/(1+9)
=-4/5

sinxcosx
=-1/2·cos(2x+π/2)
=-1/2·（-4/5）
=2/5

tan(x+π/4)=(1+tanx)/(1-tanx)=2;
1+tanx=2-2tanx;
3tanx=1;
tanx=1/3;
1+3sinxcosx-2cos²x=3/2sin2x-cos2x=(3/2)2tanx/(1+tan²x)-(1-tan²x)/(1+tan²x)
=(3/2)(2/3)/(1+1/9)-(1-1/9)/(1+1/9)
=1/9/(10/9)
=1/10;

tan(x+π/4)=3a∴(tanx+tanπ/4)/(1-tanxtanπ/4)=3(tanx+1)/(1-tanx)=3∴tanx+1=3-3tanx∴tanx=1/2sinxcosx=sinxcosx/(sin²x+cos²x)=tanx/(tan²x+1)=2/5