已知tan(π/12+α)=根号2,tan(β-π/3)=2根号2.求tan(α+β-π/4)和tan(α+β)的值
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/23 07:33:37
已知tan(π/12+α)=根号2,tan(β-π/3)=2根号2.求tan(α+β-π/4)和tan(α+β)的值
/>(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)]
从而
[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
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)]
从而
[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
已知tan(π/12+α)=根号2,tan(β-π/3)=2根号2.求tan(α+β-π/4)和tan(α+β)的值
已知tan(π/12+α)=根号2,tan(β-π/3)=2根号2,求tan(α+β)的值
已知tan[(α+β)/2]=(根号6)/2,tanαtanβ=13/7 求cos(α-β的值
tanα+tanβ+根号3tanα*tanβ=根号3 求 tan(α+β)=?
已知α+β=π/3,且α和β都是锐角,则tanα+tanβ+根号3tanαtanβ=?
若tanα=根号3(1+a),根号3(tanαtanβ+a)+tanβ=0,α,β属于(0,π/2),则α+β等于
已知tanα*tanβ=7/3 tan[(α+β)/2] = 根号2/2 求cos(α-β)
tanα=2则tan(α-π/6)=5根号3-8,
α+β=π/3,求(1+根号3tanα)(1+根号3tanβ)
由已知有tanα+tanβ=-33tanα•tanβ=4,…(2分)∴tan(α
已知tanα-tanβ>0,且tanα,tanβ是方程3x^2+5x-2=0的两个根,求tan(α+β)的值
已知α+β=60°且tanα,tanβ都存在tanα+tanβ+根号3tanαtanβ=?