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已知α为锐角，且cos[α+(π/4)]=√5/5
所以，sin[α+(π/4)]=2√5/5
则，tan[α+(π/4)]=sin[α+(π/4)]/cos[α+(π/4)]=2
==> [tanα+tan(π/4)]/[1-tanαtan(π/4)]=2
==> (tanα+1)/(1-tanα)=2
==> tanα+1=2-2tanα
==> tanα=1/3
则，sin2α=2tanα/(1+tan²α)=(2/3)/[1+(1/9)]=(2/3)/(10/9)=3/5
则，cos2α=4/5
所以，sin[2α+(π/3)]=sinαcos(π/3)+cosαsin(π/3)
=(3/5)·(1/2)+(4/5)·(√3/2)
=(3+4√3)/10
所以，sin[α+(π/4)]=2√5/5
则，tan[α+(π/4)]=sin[α+(π/4)]/cos[α+(π/4)]=2
==> [tanα+tan(π/4)]/[1-tanαtan(π/4)]=2
==> (tanα+1)/(1-tanα)=2
==> tanα+1=2-2tanα
==> tanα=1/3
则，sin2α=2tanα/(1+tan²α)=(2/3)/[1+(1/9)]=(2/3)/(10/9)=3/5
则，cos2α=4/5
所以，sin[2α+(π/3)]=sinαcos(π/3)+cosαsin(π/3)
=(3/5)·(1/2)+(4/5)·(√3/2)
=(3+4√3)/10
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