数学题在线解答已知tana+cota=5/2,a∈（π/4,π/2),求cos2a和sin(2a+π/4)的值
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数学题在线解答已知tana+cota=5/2,a∈（π/4,π/2),求cos2a和sin(2a+π/4)的值
数学题在线解答已知tana+cota=5/2,a∈（π/4,π/2),求cos2a和sin(2a+π/4)的值
数学题在线解答已知tana+cota=5/2,a∈（π/4,π/2),求cos2a和sin(2a+π/4)的值
sina/cosa + cosa/sina = 5/2
通分得
1/(cosasina) = 5/2
2/sin2a = 5/2
sin2a = 4/5
由a∈（π/4,π/2）得2a∈（π/2,π）,所以cos2a = -根号下(1-sin2a的平方) = -3/5
sin(2a+π/4) = sin2acosπ/4 + cos2asinπ/4 = 根号2/10
tana+cota
=sina/cosa+cosa/sina
=[(sina)^2+(cosa)^2]/sinacosa
=1/(1/2sin2a)=5/2
sin2a=4/5
a∈（π/4,π/2)
2a∈（π/2,π)
cos2a=-3/5
sin(2a+π/4)=sin2acosπ/4+sinπ/4cos2a=√2/10
tana+cota=5/2
tana+1/tana=5/2
因为a∈（π/4,π/2)，
tana=2
设tana=t
cos2a=(1-t^2)/(1+t^2)=-3/5
sin(α+β)=sin αcos β+cos αsin β
sin(2a+π/4)=（sin2a+cos2a）*根号2/2=(4/5-3/5)*根号2/2=根号2/10
易知，tana=2,
[1]
cos2a=(cos²a-sin²a)/(sin²a+cos²a)=(1-tan²a)/(1+tan²a)=-3/5
[2]
sin2a=(2sinacosa)/(sin²a+cos²a)=(2tana)/(1+tan²a)=4/5
∴sin[2a+(π/4)]=(√2/2)[sin2a+cos2a]=(√2/2)[(4/5)-(3/5)]=(√2)/10
tana+cota=5/2
tana+1/tana = 5/2
2(tana)^2 - 5tana +2 =0
(2tana-1)(tana-2) =0
tana=1/2( rejected)
or tana =2
sina = 2√5/5 , cosa=√5/5
cos2a = (cosa)^2 -(sina)^2 = 1/5 - 4/5...
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tana+cota=5/2
tana+1/tana = 5/2
2(tana)^2 - 5tana +2 =0
(2tana-1)(tana-2) =0
tana=1/2( rejected)
or tana =2
sina = 2√5/5 , cosa=√5/5
cos2a = (cosa)^2 -(sina)^2 = 1/5 - 4/5 = -3/5
sin(2a+π/4)
= (√2/2)(sin2a + cos2a)
= (√2/2)(2sinacosa + cos2a)
= (√2/2)(4/5-3/5)
= √2/10
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