已知a属于(0,派/2),tan(派/4+a)=2cos2a求a
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已知a属于(0,派/2),tan(派/4+a)=2cos2a求a
已知a属于(0,派/2),tan(派/4+a)=2cos2a求a
已知a属于(0,派/2),tan(派/4+a)=2cos2a求a
tan(π/4+α)=2cos2α
=> (tanπ/4+tanα)/(1-tanπ/4*tanα)= 2 (1-tan²α)/(1+tan²α)
=>(1+tanα)/(1-tanα)= 2 (1-tan²α)/(1+tan²α)
记 tanα =x,则原式=(1+x)/(1-x) =2(1-x²)/(1+x²)
里边同时约去1+x,整理得
x²-4x+1=0 ,解得 x=2±√3
又tan(π/4+α)=(1+tanα)/(1-tanα)=(1+x)/(1-x) =2cos(2α)