已知α,β是锐角,α+β≠π/2,且满足3sinβ=sin(2α+β)(1)求证:tan(α+β)=2tanα(2)求证:tanβ≤√2/4,并求等号成立时tanα与tanβ的值悬赏分可以加
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![已知α,β是锐角,α+β≠π/2,且满足3sinβ=sin(2α+β)(1)求证:tan(α+β)=2tanα(2)求证:tanβ≤√2/4,并求等号成立时tanα与tanβ的值悬赏分可以加](/uploads/image/z/12582728-8-8.jpg?t=%E5%B7%B2%E7%9F%A5%CE%B1%2C%CE%B2%E6%98%AF%E9%94%90%E8%A7%92%2C%CE%B1%2B%CE%B2%E2%89%A0%CF%80%2F2%2C%E4%B8%94%E6%BB%A1%E8%B6%B33sin%CE%B2%3Dsin%282%CE%B1%2B%CE%B2%EF%BC%89%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9Atan%28%CE%B1%2B%CE%B2%EF%BC%89%3D2tan%CE%B1%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9Atan%CE%B2%E2%89%A4%E2%88%9A2%2F4%2C%E5%B9%B6%E6%B1%82%E7%AD%89%E5%8F%B7%E6%88%90%E7%AB%8B%E6%97%B6tan%CE%B1%E4%B8%8Etan%CE%B2%E7%9A%84%E5%80%BC%E6%82%AC%E8%B5%8F%E5%88%86%E5%8F%AF%E4%BB%A5%E5%8A%A0)
已知α,β是锐角,α+β≠π/2,且满足3sinβ=sin(2α+β)(1)求证:tan(α+β)=2tanα(2)求证:tanβ≤√2/4,并求等号成立时tanα与tanβ的值悬赏分可以加
已知α,β是锐角,α+β≠π/2,且满足3sinβ=sin(2α+β)
(1)求证:tan(α+β)=2tanα
(2)求证:tanβ≤√2/4,并求等号成立时tanα与tanβ的值
悬赏分可以加
已知α,β是锐角,α+β≠π/2,且满足3sinβ=sin(2α+β)(1)求证:tan(α+β)=2tanα(2)求证:tanβ≤√2/4,并求等号成立时tanα与tanβ的值悬赏分可以加
证明:由3sinβ=sin(2α+β)得3sin[(α+β)-α]=sin[ (α+β)+α],即3[sin(α+β)cosα-cos(α+β)sinα]=sin(α+β)cosα+cos(α+β)sinα所以2sin(α+β)cosα=4cos(α+β)sinα,即tan(α+β)=2tanα.
α与β是锐角,α+β不等于90°
tan(α+β)=2tanα
(tanα+tanβ)/(1-tanα*tanβ)=2tanα
tanβ=tanα/(1+2tan^2α)
tanβ≤√2/4
tanα/(1+2tan^2α)≤√2/4
2√2*[tanα-(√2/2)]^2≥0
设tanα=√2/2,则sinα=1/√3,cos=√(2/3)
tanβ=tanα/(1+2tan^2α)=(√2/2)/[1+2(√2/2)^2]=√2/4
3sinβ=sin(2α+β)
3sinβ=sin(2α)*cosβ+cos(2α)*sinβ
3tanβ=2sinα*cosα+(1-2sin^α)*tanβ
tanβ=sinα*cosα/(1+sin^2α)=(1/√3)*√(2/3)/[1+(1/√3)^2]=√2/4
故tanα=√2/2,tanβ=√2/4
吧a+b当成一个整体化简条件即可!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
嘻嘻你是高一的的么
3sinβ=sin(2α+β)
3sin(α+β-α)=sin(α+α+β)
3sin(α+β)cos(a)-3cos(a+b)sinb=sin(a)cos(a+b)+sin(a+b)cosa
2sin(α+β)cos(a)=4sin(a)cos(a+b)
同时除以cosa
2sin(α+β)=4tanacos(a+b)同时除以cos(a+b)tan(α+β)=2tana