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Math Help - My Problem

  1. #1
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    May 2008
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    My Problem

    Guys can anyone help me with this problem:
    If Sin(Heata)=3 X Sin (Heata + 2 X Alpha)
    Prove That: Tan(Heata + Alpha) + 2 X Tan (Alpha) =0
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, omega2692!

    You have a new spelling for "theta"!

    There must be a simpler solution; I hope someone can find one.


    If sin(θ) = 3新in(θ+ 2α)

    prove that: .tan(θ + α) + 2暗an(α) .= .0 .[A]

    We are given: .sin(θ) . = . 3新in(θ + 2α) . = . 3新in(θ)搾os(2α) + 3搾os(θ)新in(α)

    Then: . sin(θ) - 3新in(θ)搾os(2α) . = . 3搾os(θ)新in(2α)

    Factor: . sin(θ) [1 - 3搾os(2α)] . = . 3搾os(θ)新in(2α)

    . . . . . . . . . . . sin(θ) . . . . . .3新in(2α)
    and we have: . -------- . = . --------------- . = . tan(θ)
    . . . . . . . . . . . cos(θ) . . . .1 - 3搾os(2α)


    . . . . . . . . . . . . . . 3[2新in(α)搾os(α)] . . . . 3新in(α).cos(α)
    Then: . tan(θ) . = . ---------------------- . = . ----------------- .[1]
    . . . . . . . . . . . . . .1 - 3[2cos(α) - 1] . . . . . 2 - 3cos(α)



    . . . . . . . . . . . . . . . . . . . . . . . tan(θ) + tan(α)
    We know that: . tan(θ + α) .= . -------------------- .[2]
    . . . . . . . . . . . . . . . . . . . . . . .1 - tan(θ)暗an(α)


    Substitute [1] into [2]:

    . . . . . . . . . . . . . . .3新in(α)搾os(α) . sin(α)
    . . . . . . . . . . . . . . ----------------- + --------
    . . . . . . . . . . . . . . . 2 - 3搾osα . . . cos(α)
    . . tan(θ + α) . = . ---------------------------------
    . . . . . . . . . . . . . . . . .3新in(α)搾os(α) .sin(α)
    . . . . . . . . . . . . . .1 + ------------------.-------
    . . . . . . . . . . . . . . . . . .2 - 3搾os(α) . .cos(α)


    . . . . . . . . . . . . . . . 3新in(α).cos(α) + 2新in(α) - 3新in(α)搾os(α)
    which simplifies to: . ----------------------------------------------------
    . . . . . . . . . . . . . . . . . 2搾os(α) - 3搾os(α) - 3新in(α)搾os(α)


    . . . . . . . . . . . . . . . . . . . 2新in(α)
    and further simplifies to: . ----------- . = . -2暗an(α)
    . . . . . . . . . . . . . . . . . . . - cos(α)



    Substitute into [A]: . -2暗an(α) + 2暗an(α) . = . 0


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