求救~sinβ=msin(2α+β)sinβ=msin(2α+β),且m≠1,α≠kπ/2,α+β≠π/2+kπ(k∈Z),求证tan(α+β)=(1+m)/(1-m)·tanα
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![求救~sinβ=msin(2α+β)sinβ=msin(2α+β),且m≠1,α≠kπ/2,α+β≠π/2+kπ(k∈Z),求证tan(α+β)=(1+m)/(1-m)·tanα](/uploads/image/z/5271058-10-8.jpg?t=%E6%B1%82%E6%95%91%7Esin%CE%B2%3Dmsin%282%CE%B1%2B%CE%B2%EF%BC%89sin%CE%B2%3Dmsin%282%CE%B1%2B%CE%B2%EF%BC%89%2C%E4%B8%94m%E2%89%A01%2C%CE%B1%E2%89%A0k%CF%80%2F2%2C%CE%B1%2B%CE%B2%E2%89%A0%CF%80%2F2%2Bk%CF%80%EF%BC%88k%E2%88%88Z%EF%BC%89%2C%E6%B1%82%E8%AF%81tan%28%CE%B1%2B%CE%B2%29%3D%EF%BC%881%2Bm%EF%BC%89%2F%281-m%29%C2%B7tan%CE%B1)
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求救~sinβ=msin(2α+β)sinβ=msin(2α+β),且m≠1,α≠kπ/2,α+β≠π/2+kπ(k∈Z),求证tan(α+β)=(1+m)/(1-m)·tanα
求救~sinβ=msin(2α+β)
sinβ=msin(2α+β),且m≠1,α≠kπ/2,α+β≠π/2+kπ(k∈Z),求证
tan(α+β)=(1+m)/(1-m)·tanα
求救~sinβ=msin(2α+β)sinβ=msin(2α+β),且m≠1,α≠kπ/2,α+β≠π/2+kπ(k∈Z),求证tan(α+β)=(1+m)/(1-m)·tanα
用a和b代替
sin[(a+b)-a]=msin[(a+b)+a]
sin(a+b)cosa-cos(a+b)sina=m[sin(a+b)cosa+cos(a+b)sina]
sin(a+b)cosa-cos(a+b)sina=msin(a+b)cosa+mcos(a+b)sina
(1-m)sin(a+b)cosa=(1+m)cos(a+b)sina
sin(a+b)cosa/cos(a+b)sina=(1+m)/(1-m)
sin(a+b)/cos(a+b)=(1+m)/(1-m)*(sina/cosa)
tan(a+b)
=(1+m)/(1-m)*tana