Question

已知tan(x+π/4)=2,则1+3sinxcosx-2cos^x=

Answer 1

已知tan(x+π/4)=(tanx+1)/(1-tanx)=2
解得tanx=1/3

1+3sinxcosx-2cos²x=sin²x+cos²x+3sinxcosx-2cos²x
=sin²x+3sinxcosx-cos²x
=(sin²x+3sinxcosx-cos²x)/(sin²x+cos²x)
同时除以cos²x，得
（tan²+3tanx-1）/ （tan²x+1）=1/10

Answer 2

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;