sinа=msin(2α+β),求证,tan(α+β)=(1+m)/(1-m)×tanα
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![sinа=msin(2α+β),求证,tan(α+β)=(1+m)/(1-m)×tanα](/uploads/image/z/5189518-46-8.jpg?t=sin%D0%B0%EF%BC%9Dmsin%282%CE%B1%2B%CE%B2%29%2C%E6%B1%82%E8%AF%81%2Ctan%28%CE%B1%2B%CE%B2%29%3D%281%2Bm%29%EF%BC%8F%281%EF%BC%8Dm%29%C3%97tan%CE%B1)
sinа=msin(2α+β),求证,tan(α+β)=(1+m)/(1-m)×tanα
sinа=msin(2α+β),求证,tan(α+β)=(1+m)/(1-m)×tanα
sinа=msin(2α+β),求证,tan(α+β)=(1+m)/(1-m)×tanα
sinb=msin(2a+b)
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]
(1-m)sin(a+b)cosa=(m+1)cos(a+b)sina
tan(a+b)=(1+m)/(1-m)tana