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∵α∈(0,π/2),∴α+π/4∈(π/4,3π/4)
∵cos(α+π/4)=1/3，
∴sin(α+π/4)=√[1-cos²(α+π/4)]=2√2/3
∴cosα=cos[(α+π/4)-π/4]
=cos(α+π/4)cosπ/4+sin(α+π/4)sinπ/4
=1/3*√2/2+2√2/3*√2/2
=(√2+4)/6
∵cos(α+π/4)=1/3，
∴sin(α+π/4)=√[1-cos²(α+π/4)]=2√2/3
∴cosα=cos[(α+π/4)-π/4]
=cos(α+π/4)cosπ/4+sin(α+π/4)sinπ/4
=1/3*√2/2+2√2/3*√2/2
=(√2+4)/6
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