第一题怎么做?求解答
2个回答
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答案为C
tan(α+β)=(tanα+tanβ)/(1-tanα tanβ)
tan(α+π/3)=(tanα+tanπ/3)/(1-tanα tanπ/3)
(tanα+√3)/(1-√3tanα )=4/3
4-4√3tanα=3tanα +3√3
tanα=(4-3√3)/(3+4√3)
tan²α=(43-24√3)/(57+24√3)
万能公式:
1+tan²α=1/cos²α
α∈(π/2,π)
cosα<0
cosα=-(3+4√3)/10
tan(α+β)=(tanα+tanβ)/(1-tanα tanβ)
tan(α+π/3)=(tanα+tanπ/3)/(1-tanα tanπ/3)
(tanα+√3)/(1-√3tanα )=4/3
4-4√3tanα=3tanα +3√3
tanα=(4-3√3)/(3+4√3)
tan²α=(43-24√3)/(57+24√3)
万能公式:
1+tan²α=1/cos²α
α∈(π/2,π)
cosα<0
cosα=-(3+4√3)/10
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