三角學的問題

已知x、y為銳角，cosx=4/5，tan(x-y)=(-1/3)，求cosx
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已知x、y為銳角，cosx=4/5，tan(x-y)=(-1/3)，求cosy
cosx = 4/5
sinx = 3/5
tan(x - y) = -1/3
sin(x - y)/cos(x - y) = -1/3
3sinycosx - 3sinxcosy = cosxcosy + sinxsiny
3sinycosx - sinxsiny = cosxcosy + 3sinxcosy
cosy(cosx + 3sinx) = 3sinycosx - sinxsiny
cosy(cosx + 3sinx) = siny(3cosx - sinx)
tany = (cosx + 3sinx) / (3cosx - sinx)
tany = [(4/5) + (9/5)] / [(12/5) - (3/5)]
tany = (13/5) / (9/5)
tany = 13/9
cosy = 9/[5(10)^1/2]

??cosx=4/5(given)
你係唔係話cos y
cos x=4/5
x=36.86989765.....
tan(x-y)=(-1/3)
x-y= -18.43494882
y=55.0484647