已知cosа=1/7.cos(а-β)=13/14.且0<β<а<∏/2.求tan2а已知cosа=1/7.cos(а-β)=13/14.且0<β<а<∏/2.求tan2а和β
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![已知cosа=1/7.cos(а-β)=13/14.且0<β<а<∏/2.求tan2а已知cosа=1/7.cos(а-β)=13/14.且0<β<а<∏/2.求tan2а和β](/uploads/image/z/2498039-71-9.jpg?t=%E5%B7%B2%E7%9F%A5cos%D0%B0%EF%BC%9D1%EF%BC%8F7.cos%EF%BC%88%D0%B0%EF%BC%8D%CE%B2%EF%BC%89%EF%BC%9D13%EF%BC%8F14.%E4%B8%940%EF%BC%9C%CE%B2%EF%BC%9C%D0%B0%EF%BC%9C%E2%88%8F%EF%BC%8F2.%E6%B1%82tan2%D0%B0%E5%B7%B2%E7%9F%A5cos%D0%B0%EF%BC%9D1%EF%BC%8F7.cos%EF%BC%88%D0%B0%EF%BC%8D%CE%B2%EF%BC%89%EF%BC%9D13%EF%BC%8F14.%E4%B8%940%EF%BC%9C%CE%B2%EF%BC%9C%D0%B0%EF%BC%9C%E2%88%8F%EF%BC%8F2.%E6%B1%82tan2%D0%B0%E5%92%8C%CE%B2)
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已知cosа=1/7.cos(а-β)=13/14.且0<β<а<∏/2.求tan2а已知cosа=1/7.cos(а-β)=13/14.且0<β<а<∏/2.求tan2а和β
已知cosа=1/7.cos(а-β)=13/14.且0<β<а<∏/2.求tan2а
已知cosа=1/7.cos(а-β)=13/14.且0<β<а<∏/2.求tan2а和β
已知cosа=1/7.cos(а-β)=13/14.且0<β<а<∏/2.求tan2а已知cosа=1/7.cos(а-β)=13/14.且0<β<а<∏/2.求tan2а和β
sin^2α=48/49
tan2α=sin2α/cos2α=-8√3/47
cosβ=cos(α-(α-β))=cosαcos(α-β)+sinαsin(α-β)=1/2
β=π/3