Question

若a属于(π/2,π),tan(a+π/4)=1/7,则sin a为

Answer 1

tan(a+π/4)=(tana+1)/(1-tana)=1/7
7tana+7=1-tana
8tana=-6
tana=-3/4
tana=sina/cosa=-3/4
则：cosa=-4sina/3
代入：sin²a+cos²a=1得：
25sin²a/9=1
得：sin²a=9/25
因为a属于(π/2,π),
所以，sina>0
所以，sina=3/5

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Answer 2

tan(a+Pai/4)=(tana+tanPai/4)/(1-tanatanPai/4)=(tana+1)/(1-tana)=1/7
7+7tana=1-tana
tana=-3/4
sina/cosa=-3/4
cosa=-4/3sina
(sina)^2+(cosa)^2=1
(sina)^2+16/9*(sina)^2=1
(sina)^2=9/25
sina>0
sina=3/5

Answer 3

tan(a+π/4)=(tana+tanπ/4)/(1-tanatanπ/4)
=(1+tana)/(1-tana)=1/7
∴7+7tana=1-tana
tana=-3/4
∴sina=3/√(3²+4²)=3/5

Answer 4

tan(a+π/4)=(tana+tanπ/4)/(1-tanatanπ/4)=1/7

7(tana+1)=1-tana
tana=-3/4
∵a∈(π/2,π)
∴sina=-tana/√(tan²a+1)
=3/4/√[(-3/4)²+1]
=3/5