已知tan[α+π/4]=2,求cos2α+3sin²α的值

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已知tan[α+π/4]=2,求cos2α+3sin²α的值

已知tan[α+π/4]=2,求cos2α+3sin²α的值
已知tan[α+π/4]=2,求cos2α+3sin²α的值

已知tan[α+π/4]=2,求cos2α+3sin²α的值
tan(α+π/4)=(tanα+tanπ/4)/(1-tanα·tanπ/4) =2
(tanα+1)/(1-tanα)=2
2-2tanα=tanα+1
tanα=1/3
cosα=3sinα
sin²α+cos²α=1
10sin²α=1
sin²α=1/10
cos2α+3sin²α=1-2sin²α+3sin²α=1+sin²α=1+1/10=11/10

化简得:cos2α+3sin²α=2tan²α+1
tan[α+π/4]=(tanα+tanπ/4)/(1-tanαtanπ/4)=2
解得tan α=1/3

cos2α+3sin²α=11/9