sin(π/4+α)=5/13 α∈(π/4,3π/4) 求(1+tanα)/(1-tanα)如题
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sin(π/4+α)=5/13α∈(π/4,3π/4)求(1+tanα)/(1-tanα)如题sin(π/4+α)=5/13α∈(π/4,3π/4)求(1+tanα)/(1-tanα)如题sin(π/
sin(π/4+α)=5/13 α∈(π/4,3π/4) 求(1+tanα)/(1-tanα)如题
sin(π/4+α)=5/13 α∈(π/4,3π/4) 求(1+tanα)/(1-tanα)
如题
sin(π/4+α)=5/13 α∈(π/4,3π/4) 求(1+tanα)/(1-tanα)如题
因为 α ∈(π/4,3π/4),
所以 π/4 +α ∈(π/2,π),
又因为 sin (π/4 +α) =5/13,
所以 cos (π/4 +α) = -根号[ 1 -(5/13)^2 ]
= -12/13.
所以 tan (π/4 +α) =sin (π/4 +α) /cos (π/4 +α)
= -5/12.
又因为 tan (π/4 +α) =(tan π/4 +tan α) /(1 -tan π/4 tan α)
=(1 +tan α) /(1 -tan α),
所以 (1 +tan α) /(1 -tan α) = -5/12.
= = = = = = = = =
以上计算可能有误.
注意 tan π/4 =1,因此
(1 +tan α) /(1 -tan α) =tan (π/4 +α).
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