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=( )

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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=( )

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=( )
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