已知sin 2α=5/13,π/4<α<π/2,

解:由于π/4<α<π/2
则π/2<2α<π
则有: sin2α>0,cos2α<0
则: tan2α=sin2α/cos2α<0
则:cos2α=-根号[1-sin^2(2α)]
=-12/13
则sin 4α=2sin2αcos2α
=-120/13
cos 4α=2cos^2(2α)-1
=119/169
tan 4α=sin 4α/cos 4α
=-1560/119

sin 2α=5/13, π/4<α<π/2,
π/2<2α<π(第二象限,余弦值为负）
cos2α=-12/13
sin4α=2sin2αcos2α=-120/169
cos4α=(cos2α)^2-(sin2α)^2=119/169
tan 4α=sin4α/cos4α=-120/119

由π/4<α<π/2可得π/2<2α<π，cos 2α=-12/13
sin 4α=2sin 2α*cos 2α=2*（5/13）*（-12/13）=-120/169
cos 4α=1-2sin 2α*sin 2α=1-2（5/13）*（5/13）=119/169
tan 4α=sin 4α/cos 4α=（-120/169）/（119/169）=-120/119