一道数学证明题已知锐角α,β,满足tan(α-β)=sin2β,求证tanα+tanβ=2tan2β
2019-05-27
一道数学证明题
已知锐角α,β,满足tan(α-β)=sin2β,求证tanα+tanβ=2tan2β
优质解答
sin2β
= 2tanβ/(1+tan^2 β)
= [2tanβ*(1+2tan^2 β)]/[(1+tan^2 β)^2]
= (2tanβ+2tan^3 β)/(1+2tan^2 β + tan^4 β )
= [(2tanβ+2tan^3 β)/(1-tan^2 β)] /
[(1-tan^2 β + 3tan^2 β + tan^4 β )/(1-tan^2 β)]
= [(3tanβ+tan^3 β)/(1-tan^2 β) - tanβ] /
[1+(3tanβ+tan^3 β)/(1-tan^2 β) * tanβ]
=tan(α-β)
= (tanα - tanβ)/(1+tanα * tanβ)
所以 tanα=(3tanβ+tan^3 β)/(1-tan^2 β)
= 4tanβ/(1-tan^2 β)-tanβ
=2tan2β-tanβ
因此tanα+tanβ=2tan2β
sin2β
= 2tanβ/(1+tan^2 β)
= [2tanβ*(1+2tan^2 β)]/[(1+tan^2 β)^2]
= (2tanβ+2tan^3 β)/(1+2tan^2 β + tan^4 β )
= [(2tanβ+2tan^3 β)/(1-tan^2 β)] /
[(1-tan^2 β + 3tan^2 β + tan^4 β )/(1-tan^2 β)]
= [(3tanβ+tan^3 β)/(1-tan^2 β) - tanβ] /
[1+(3tanβ+tan^3 β)/(1-tan^2 β) * tanβ]
=tan(α-β)
= (tanα - tanβ)/(1+tanα * tanβ)
所以 tanα=(3tanβ+tan^3 β)/(1-tan^2 β)
= 4tanβ/(1-tan^2 β)-tanβ
=2tan2β-tanβ
因此tanα+tanβ=2tan2β