①定义域:令sin(x+π/2)≠0,解得:x≠(π/2)+kπ(k∈R)
②∵cos2α=2cos²α-1
∴cos2α= -7/25
又∵∠a在第一象限,且cosα=3/5
∴sinα=4/5
∴sin2α=2·sinα·cosα=24/25
f(α)=[1+√2cos(2α-π/4)]/sin(α+π/2)
=[1+√2(cos2α·cos(π/4)+sin2α·sin(π/4)]/cosα
=14/5
①定义域:令sin(x+π/2)≠0,解得:x≠(π/2)+kπ(k∈R)
②∵cos2α=2cos²α-1
∴cos2α= -7/25
又∵∠a在第一象限,且cosα=3/5
∴sinα=4/5
∴sin2α=2·sinα·cosα=24/25
f(α)=[1+√2cos(2α-π/4)]/sin(α+π/2)
=[1+√2(cos2α·cos(π/4)+sin2α·sin(π/4)]/cosα
=14/5