已知tanα+1/tanα=5/2,α属于(π/4,π/2),求cos2α和sin(2α+π/4)

tanα+1/tanα=5/2可得tanα=1/2或2
而α属于(π/4,π/2）知α属于[1，+∝），所以tanα=2
cos2α={（cosα)^2-(sinα)^2}/{(sinα)^2+(cosα)^2}
={1-(tanα)^2}/{1+（tanα）^2}= -3/5
sin2α=2sinαcosα=4/5
sin(2α+π/4)=√2/2（sin2α+cos2α）= √2/10

tanα+1/tanα=5/2
sinα/cosα+cosα/sinα=5/2
(sin²α+cos²α)/sinαcosα=5/2
1/sinαcosα=5/2
sinαcosα=2/5
2sinαcosα=4/5
sin2α=4/5
α∈(π/4,π/2),
2α∈(π/2,π)
所以：cos2αcos2α=-根号(1-sin²2α)
=-根号(9/25)
=-3/5
sin(2α+π/4)
=sin2αcosπ/4+cos2αsinπ/4
=√2/2×(4/5-3/5)
=√2/10

∵α∈(π/4,π/2) ∴tanα＞1∵tanα＋1/tanα＝5/2＝2＋1/2 ∴tanα＝2∴cos2α＝(1－tan²α)/(1＋tan²α)＝﹣3/5 sin2α＝2tanα/(1＋tan²α)＝4/5 sin(2α＋π/4)＝√2/2(sin2...