Question

tan(a+π/4)=1/2,-π/2<a<0,则2sin^a+sin2a/cos(a-π/4)

Answer 1

∵tan(a+π/4)=1/2
∴(1+tana)/(1-tana)=1/2
∴2+2tana=1-tana
∴tana=-1/3
∴sina/cosa=-1/3
∴cosa=-3sina代入sin²a+cos²a=1
得10sin²a=1,sin²a=1/10
∵,-π/2<a<0∴sina=-√10/10
（2sin²a+sin2a）/cos(a-π/4)
=（2sin²a+2sinacosa)/[√2/2(cosa+sina)]
=2√2(sina+cosa)sinsa/(cosa+sina)
=2√2sina
=2√2*(-√10/10)
=-2√5/5

追问

sina=√5/5,a是（0，π/2),tanB=1/3.求tana的值。求tan(a+2B）的值。

追答

刚才太忙
∵sina=√5/5,a是（0，π/2),
∴cosa=√(1-sin²a)=2√5/5
∴tana=sina/cosa=1/2
∵tanB=1/3.
∴tan2B=2tanB/(1-tan²B)
=(2/3)/(1-1/9)=3/4
∴tan(a+2B）
=(tana+tan2B)/(1-tanatan2B)
=(1/2+3/4)/(1-3/8)=2