(1)
tan(π/4+α)=[tan(π/4)+tanα]/[1-tan(π/4)*tanα]=(1+tanα)/(1-tanα)=3
1+tanα=3*(1-tanα),即4tanα=2,tanα=1/2
(2)
2sinαcosα+cos2α
=(2sinαcosα+cos2α)/[(sinα)^2+(cosα) ^2]
=[2sinαcosα+(cosα)^2-(sinα)^2]/[(sinα)^2+(cosα) ^2] 上下同除以(cosα) ^2
=[2tanα+1-(tanα)^2]/[(tanα)^2+1]
=(1+1-1/4)/(1/4+1)
=7/5