已知tan（π/12+α）=根號(hào)2,tan（β-π/3）=2根號(hào)2.求tan（α+β-π/4）和tan（α+β）的值
已知tan（π/12+α）=根號(hào)2,tan（β-π/3）=2根號(hào)2.求tan（α+β-π/4）和tan（α+β）的值
數(shù)學(xué)人氣：868 ℃時(shí)間：2020-05-23 03:17:49
優(yōu)質(zhì)解答
（1）
tan(α+β-π/4)
=tan(π/12+α+β-π/3)
=[tan(π/12+α)+tan(β-π/3)]/[1-tan(π/12+α)tan(β-π/3)]
=(√2+2√2)/[1-√2*2√2]
=3√2/(1-4)
=-√2
（2）
tan(α+β-π/4)
=[tan(α+β)-tanπ/4]/[1+tan(α+β)tanπ/4]
=[tan(α+β)-1]/[1+tan(α+β)]
從而
[tan(α+β)-1]/[1+tan(α+β)]=-√2
tan(α+β)-1=-√2[1+tan(α+β)]
tan(α+β)-1=-√2-√2tan(α+β)
tan(α+β)+√2tan(α+β)=1-√2
(1+√2)tan(α+β)=1-√2
于是tan(α+β)
=(1-√2)/(1+√2)
=(1-√2)²/(1+√2)(1-√2)
=-(1-√2)²
=-3+2√2
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