已知α=π/9,求值tanα+tan2α+根号3tanα×tan2α=( )
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/06/25 20:14:19
已知α=π/9,求值tanα+tan2α+根号3tanα×tan2α=( )
已知α=π/9,求值tanα+tan2α+根号3tanα×tan2α=( )
已知α=π/9,求值tanα+tan2α+根号3tanα×tan2α=( )
![已知α=π/9,求值tanα+tan2α+根号3tanα×tan2α=( )](/uploads/image/z/7767969-33-9.jpg?t=%E5%B7%B2%E7%9F%A5%CE%B1%3D%CF%80%2F9%2C%E6%B1%82%E5%80%BCtan%CE%B1%2Btan2%CE%B1%2B%E6%A0%B9%E5%8F%B73tan%CE%B1%C3%97tan2%CE%B1%3D%EF%BC%88+%EF%BC%89)
tan(α+2α)=tan(3α)=tan(3·π/9)=tan(π/3)=√3
又tan(α+2α)=[tanα+tan(2α)]/[1-tanα·tan(2α)]
因此[tanα+tan(2α)]/[1-tanα·tan(2α)]=√3
tanα+tan(2α)=√3-√3tanα·tan(2α)
tanα+tan(2α)+√3tanα·tan(2α)=√3
又tan(α+2α)=[tanα+tan(2α)]/[1-tanα·tan(2α)]
因此[tanα+tan(2α)]/[1-tanα·tan(2α)]=√3
tanα+tan(2α)=√3-√3tanα·tan(2α)
tanα+tan(2α)+√3tanα·tan(2α)=√3
已知α=π/9,求值tanα+tan2α+根号3tanα×tan2α=( )
已知tan(α+β)=3,tan(α-β)=5,求tan2α,tan2β的值
证明:tan(α+π/4)+tan(α+3π/4)=2tan2α
已知(1+tan2α)/(1-tanα)=2010,求(1/cos2α)+tan2α
已知tan2α=⅓,求tanα的值
化简tan2αtan(π/6-α)+tan2αtan(π/3-α)+tan(π/6-α)tan(π/3-α)=?请写下过
已知tan(α+β)=3,tan(α-β)=5,求tan2α和tan2β的值?
1.已知tan2分之α=2,求值:⑴ tan(α+4分之π);⑵ (6sinα+cosα)/(3sinα-2cosα)
求证:tan(a+π/4)+tan(a-π/4)=2tan2α
已知tan(α+β)=3,tan(α-β)=2,求2α,tan2β的值
三角函数计算问题已知tan(α+β)=3,tan(α-β)=5,求tan2α,tan2β的值
求证tan2分之α-1/tan2分之α=-tanα分之2