[2sin²α+sin(2α)]÷cos(α-π/4)
=(2sin²α+2sinαcosα)/[cosαcos(π/4)+sinαsin(π/4)]
=2(√2)sinα(sinα+cosα)/(cosα+sinα)
=2(√2)sinα
即：[2sin²α+sin(2α)]÷cos(α-π/4)=2(√2)sinα……………………(1)
tan(α+π/4)=1/2
[tanα+tan(π/4)]/[1-tanαtan(π/4)]=1/2
(tanα+1)/(1-tanα)=1/2
2(tanα+1)=1-tanα
3tanα=-1
tanα=-1/3
sinα/cosα=-1/3
3sinα=-cosα
3sinα=-√(1-sin²α)
9sin²α=1-sin²α
sin²α=1/10
sinα=±(√10)/10
因為：-π/2＜α＜0
所以：sinα=(√10)/10
代入(1),有：
[2sin²α+sin(2α)]÷cos(α-π/4)=2(√2)[(√10)/10]
=(2/5)√5
即：
[2sin²α+sin(2α)]÷cos(α-π/4)=(2/5)√5