作业帮已知tana=1/7,tanβ=1/3,求tan（a+2β）
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已知tana=1/7,tanβ=1/3,求tan（a+2β）
tan(a+β)=7
(tana+tanβ)/(1-tanatanβ) = 7
3(tana+tanβ) = 7
3sin(a+β)/cosacosβ = 7
cosacosβ= 3sin(a+β)/7
= 3(7/5√2)/7
= 3/(5√2) = 3√2/10
tanatanβ = 2/3
sinasinβ/cosacosβ = 2/3
sinasinβ = (2/3)cosacosβ = (2/3)(3√2/10)= √2/5
cos(a-β) =cosacosβ+ sinasinβ
= 3√2/10 + √2/5
= √2/2
(tana+tanβ)/(1-tanatanβ) = 7
3(tana+tanβ) = 7
3sin(a+β)/cosacosβ = 7
cosacosβ= 3sin(a+β)/7
= 3(7/5√2)/7
= 3/(5√2) = 3√2/10
tanatanβ = 2/3
sinasinβ/cosacosβ = 2/3
sinasinβ = (2/3)cosacosβ = (2/3)(3√2/10)= √2/5
cos(a-β) =cosacosβ+ sinasinβ
= 3√2/10 + √2/5
= √2/2
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