已知tanα+1/tanα=5/2,α属于(π/4,π/2),求cos2α和sin(2α+π/4)
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tanα+1/tanα=5/2
sinα/cosα+cosα/sinα=5/2
(sin²α+cos²α)/sinαcosα=5/2
1/sinαcosα=5/2
sinαcosα=2/5
2sinαcosα=4/5
sin2α=4/5
α∈(π/4,π/2),
2α∈(π/2,π)
所以:cos2α<0
cos2α=-根号(1-sin²2α)
=-根号(9/25)
=-3/5
sin(2α+π/4)
=sin2αcosπ/4+cos2αsinπ/4
=√2/2×(4/5-3/5)
=√2/10
sinα/cosα+cosα/sinα=5/2
(sin²α+cos²α)/sinαcosα=5/2
1/sinαcosα=5/2
sinαcosα=2/5
2sinαcosα=4/5
sin2α=4/5
α∈(π/4,π/2),
2α∈(π/2,π)
所以:cos2α<0
cos2α=-根号(1-sin²2α)
=-根号(9/25)
=-3/5
sin(2α+π/4)
=sin2αcosπ/4+cos2αsinπ/4
=√2/2×(4/5-3/5)
=√2/10
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2011-12-11
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tanα+1/tanα=5/2可得tanα=1/2或2
而α属于(π/4,π/2)知α属于[1,+∝),所以tanα=2
cos2α={(cosα)^2-(sinα)^2}/{(sinα)^2+(cosα)^2}
={1-(tanα)^2}/{1+(tanα)^2}= -3/5
sin2α=2sinαcosα=4/5
sin(2α+π/4)=√2/2(sin2α+cos2α)= √2/10
而α属于(π/4,π/2)知α属于[1,+∝),所以tanα=2
cos2α={(cosα)^2-(sinα)^2}/{(sinα)^2+(cosα)^2}
={1-(tanα)^2}/{1+(tanα)^2}= -3/5
sin2α=2sinαcosα=4/5
sin(2α+π/4)=√2/2(sin2α+cos2α)= √2/10
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