已知sinβ=msin(2α+β) 证明tan(α+β)=(1+m)/(1-m)tanα
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![已知sinβ=msin(2α+β) 证明tan(α+β)=(1+m)/(1-m)tanα](/uploads/image/z/5189515-43-5.jpg?t=%E5%B7%B2%E7%9F%A5sin%CE%B2%3Dmsin%282%CE%B1%2B%CE%B2%EF%BC%89+%E8%AF%81%E6%98%8Etan%28%CE%B1%2B%CE%B2%EF%BC%89%3D%281%2Bm%29%2F%EF%BC%881-m%EF%BC%89tan%CE%B1)
已知sinβ=msin(2α+β) 证明tan(α+β)=(1+m)/(1-m)tanα
已知sinβ=msin(2α+β) 证明tan(α+β)=(1+m)/(1-m)tanα
已知sinβ=msin(2α+β) 证明tan(α+β)=(1+m)/(1-m)tanα
sinβ=msin(2α+β)
sin[(a+b)-a]=msin[(a+b)+a]
sin(a+b)cosa-cos(a+b)sina=msin(a+b)cosa+mcos(a+b)sina
(1-m)sin(a+b)cosa=(m+1)cos(a+b)sina
tan(α+β)=(1+m)/(1-m)tanα