1樓:匿名使用者
(1)求tanα的值
∵tan(α+β)=1/5,tan(β+π/4)=1/4∴tan(α-π/4)=[tan(α+β)-tan(β+π/4)]/[1+tan(α+β)tan(β+π/4)]=(1/5-1/4)/(1+1/5*1/4)=-1/21
∵tan(α-π/4)=(tanα-tanπ/4)/(1+tanαtanπ/4)=(tanα-1)/(1+tanα)
∴tanα-1+(1+tanα)/21=0得tanα=10/11
(2)求sin²α+sinαcosα+cos²α的值∵sinα/cosα=tanα=10/11∴sin²α=100/121cos²α
∵sin²α+cos²α=1
∴cos²α=121/221
sin²α+sinαcosα+cos²α
=cos²α(tan²α+tanα+1)
=121/221*(100/121+10/11+1)=331/221
2樓:皮皮鬼
解,tan(β+π/4)=1/4
即(tanβ+tanπ4)/(1-tanβ*tanπ/4)=1/4解得tanβ=-3/5
由tan(α+β)=1/5
即(tan(α)+tan(β))/1-tan(α)*tan(β)=1/5
解得tan(α)=5/3
cos²α+sin²α+sinαcosα
=(cos²α+sin²α+sinαcosα)/(cos²α+sin²α)
=[(cos²α+sin²α+sinαcosα)*1/cos²α]/[(cos²α+sin²α)1/cos²α]
=(1+tan²α+tanα)/(1+tan²α)=(1+(5/3)²+(5/3))/(1+(5/3)²)=1+15/34
=49/34