第2个回答 2019-09-20
(1)解:tanα=sinα/cosα=4/3
则sinα=(4/3)cosα
∵sin²α + cos²α=1
∴[(4/3)cosα]² + cos²α=1
(16/9)cos²α + cos²α=1
(25/9)cos²α=1
cos²α=9/25,则cosα=±3/5
∵α是锐角
∴cosα=3/5
∴cos2α=2cos²α - 1
=2•(3/5)² - 1=-7/25
(2)由(1)得:sinα=(4/3)cosα
=(4/3)•(3/5)=4/5
sin2α=2sinαcosα=24/25
∵α,β是锐角,且cos(α+β)<0
∴sin(α+β)=√1-cos²(α+β)
=(2√5)/5
∴cos(α-β)=cos[2α-(α+β)]
=cos2αcos(α+β) + sin2αsin(α+β)
=(-7/25)•(-√5/5) + (24/25)•[(2√5)/5]
=(11√5)/25
sin(α-β)=sin[2α-(α+β)]
=sin2αcos(α+β) - cos2αsin(α+β)
=(24/25)•(-√5/5) - (-7/25)•[(2√5)/5]
=-(2√5)/25
∴tan(α-β)=sin(α-β)/cos(α-β)
=[-(2√5)/25]/[(11√5)/25]
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