0 < α≤π/2 -π/2 ≤ - α < 0 -π/4 ≤π/4 - α < π/4 又sin(π/4 - α) = 5/13 > 0,∴0 ≤π/4 - α < π/4 sin(π/4)cosα - cos(π/4)sinα = 5/13 (√2/2)cosα - (√2/2)sinα = 5/13 cos2α/cos(π/4 + α) = (1 - 2sinα)/[cos(π/4)cosα - sin(π/4)sinα] = [1 - 2sin((π/4 - α) - π/4)]/[(√2/2)cosα - (√2/2)sinα] = {1 - 2[sin(π/4 - α)cos(π/4) - cos(π/4 - α)sin(π/4)]}/(5/13) = {1 - 2[(5/13)(√2/2) - (12/13)(√2/2)]}/(5/13) = (120/169)/(5/13) = 24/13
sin(π/4-a)=5/13,且a∈(0,π/2),则cos2a/cos(π/4+a)的值为?
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