Question

已知tan(π/4+α）=2，求1/2sinαcosα+（cosα）^2

Answer 1

tan(π/4+α）=2
(tanπ/4+tanα)/(1-tanπ/4tanα)=2
(1+tanα)/(1-tanα)=2
1+tanα=2(1-tanα)
1+tanα=2-2tanα
3tanα=1
tanα=1/3

1/2sinαcosα+（cosα）^2
=1/4sin2α+(1+cos2α)/2
=1/4sin2α+1/2*cos2α+1/2
=1/4*2tanα/(1+tan^2α)+1/2*(1-tan^2α)/(1+tan^2α)+1/2
=1/2*tanα/(1+tan^2α)+1/2*(1-tan^2α)/(1+tan^2α)+1/2
=(1/2*1/3)/[1+(1/3)^2]+1/2*[1-(1/3)^2]/[1+(1/3)^2]+1/2
=(1/6)/(10/9)+(4/9)/(10/9)+1/2
=3/20+4/10+1/2
=3/20+8/20+10/20
=21/20