1/tanα+1/cotα=5/2,求2sin*2(3π-α）-3cos(π/2+α)cos(3π/2-α)+2的值

1/tanα+1/cotα=5/2,求2sin*2(3π-α）-3cos(π/2+α)cos(3π/2-α)+2的值
1/tanα+1/cotα=5/2,求2sin*2(3π-α）-3cos(π/2+α)cos(3π/2-α)+2的值

1/tanα+1/cotα=5/2,求2sin*2(3π-α）-3cos(π/2+α)cos(3π/2-α)+2的值
1/tanα+1/cotα=5/2
tanα=2或1/2
sin*2(3π-α）-3cos(π/2+α)cos(3π/2-α)+2
=2sin*2(α）-3cos(α)sin(-α)+2
=[2sin*2(α）+3cos(α)sin(α)]/(sin*2（α）+cos*2(α）) + 2
=[2tan*2(α）+3tan(α）]/(tan*2（α）+1) +2
当tanα=2时,原式= 24/5
当tanα=1/2时,原式=18/5

1/tanα+1/cotα=5/2
cosα/sinα+sinα/cosα=5/2
(sin²α+cos²α)/sinαcosα=5/2
sinαcosα=2/5
sin2α=4/5
cos2α=±√(1-sin²2α)=±3/5
所以2sin*2(3π-α）-3cos(π/2+α)cos(3π/2-α)+2
=...

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1/tanα+1/cotα=5/2
cosα/sinα+sinα/cosα=5/2
(sin²α+cos²α)/sinαcosα=5/2
sinαcosα=2/5
sin2α=4/5
cos2α=±√(1-sin²2α)=±3/5
所以2sin*2(3π-α）-3cos(π/2+α)cos(3π/2-α)+2
=2sin²α-3(-sinα)(-sinα)+2
=2sin²α-3sin²α+2
=2-sin²α
=1-(1-cos2α)/2
=(1+cos2α)/2
=(1+3/5)/2=4/5
或=(1-3/5)/2=1/5

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化简得2-sin^2(α)，
有tanα=2或1/2，
可得结果=6/5,9/5