Question

已知sin(α-π/4)=(7根号2)/10.且π/2<α<(3π)/4.求tan(2α+π/4)

Answer 1

sin(α-π/4)=(7√2)/10
π/2<α<(3π)/4.
π/4<α－π/4<π/2.
cos(α-π/4)=√2/10
sinα-cosα=7/5........(1)
sinα+cosα=1/5........(2)
解（1）（2）得
sinα＝4/5, cosα=-3/5
tanα=sinα/cosα=-4/3
tan2α=2tanα/[1-(tanα)^2]=2*(-4/3)/[1-16/9]=24/7
tan(2α+π/4)=(1+tan2α)/(1-tan2α)=(1+24/7)/(1-24/7)=-31/17
tan(2α+π/4)=-31/17

Answer 2

π/2<α<(3π)/4.
→π/4<α-π/4<π/2.
则cos(α-π/4)=√[1-sin^2 (α-π/4)]=√2/10.

sin(2α-π/2)
=2·sin(α-π/4)·cos(α-π/4)
=7/25.
即cos 2α=-7/25
∴sin 2α
=√[1-cos^2 2α]
=-24/25.

则tan 2α=sin 2α /cos 2α=24/7.
则tan(2α+π/4)
=(tan 2α + tan π/4) /(1- tan 2α · tan π/4)
=(24/7 + 1)/[1-(24/7)×1]
= -31/17