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Math Help - Addition Formulae for Trigonometry

  1. #1
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    Addition Formulae for Trigonometry



    I need someone to check if part a) is correct and need help on part b)
    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by r_maths View Post


    I need someone to check if part a) is correct and need help on part b)
    Thanks
    Here's (a). your logic is correct, however, you can't call both bases adj, since they represend different lengths, call one adj1 and the other adj2, i used different names though
    Attached Thumbnails Attached Thumbnails Addition Formulae for Trigonometry-solution.gif  
    Last edited by Jhevon; April 17th 2007 at 11:29 AM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by r_maths View Post


    I need someone to check if part a) is correct and need help on part b)
    Thanks
    Here's (b)

    Darn it, I forgot to say "this is my 12th post" last time!
    Attached Thumbnails Attached Thumbnails Addition Formulae for Trigonometry-solution.gif  
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  4. #4
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    Thanks for the speedy response Jhevon!
    Congrats on 1200
    I'm privileged.
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  5. #5
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    Hello, r_maths!

    You're doing fine!
    Code:
                  C
                  *
                 *| *
                * |   *
               *  |     *
              *   |h      *
             *    |         *
            *     |           *
           * α    |           β *
        A * * * * * * * * * * * * * B
                  D

    In right triangle CDA: .tanα = h/AD . . AD = h/tanα

    In right triangle CDB: .tanβ = h/DB . . DB = h/tanβ

    . . . . . . . . . . . . . . . . . . . . h . . . . h
    Then: .AB .= .AD + DB .= .------ + ------
    . . . . . . . . . . . . . . . . . . . tanα . . tanβ

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    For part (b), we have:

    . . . . . . . . h . . . . .h . . . . h搾osβ . . h搾osα
    . . AB .= .------ + ------ .= .-------- + --------
    . . . . . . . tanβ . . tanα . . . .sinβ . . . . sinα


    . . . . . . .h新inα搾osβ + h新inβ搾osα . . . .h(sinα搾osβ + sinB搾osα)
    . . . . = .-------------------------------- .= .------------------------------
    . . . . . . . . . . . . sinα新inβ . . . . . . . . . . . . . . sinα新inβ


    . . . . . . .h新in(α + β)
    . . . . = .---------------
    . . . . . . . .sinα新inβ



    Edit:Too slow . . again
    Last edited by Soroban; April 17th 2007 at 11:55 AM.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, r_maths!

    You're doing fine!
    Code:
                  C
                  *
                 *| *
                * |   *
               *  |     *
              *   |h      *
             *    |         *
            *     |           *
           * α    |           β *
        A * * * * * * * * * * * * * B
                  D

    In right triangle CDA: .tanα = h/AD . . AD = h/tanα

    In right triangle CDB: .tanβ = h/DB . . DB = h/tanβ

    . . . . . . . . . . . . . . . . . . . . h . . . . h
    Then: .AB .= .AD + DB .= .------ + ------
    . . . . . . . . . . . . . . . . . . . tanα . . tanβ

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    For part (b), we have:

    . . . . . . . . h . . . . .h . . . . h搾osβ . . h搾osα
    . . AB .= .------ + ------ .= .-------- + --------
    . . . . . . . tanβ . . tanα . . . .sinβ . . . . sinα


    . . . . . . .h新inα搾osβ + h新inβ搾osα . . . .h(sinα搾osβ + sinB搾osα)
    . . . . = .-------------------------------- .= .------------------------------
    . . . . . . . . . . . . sinα新inβ . . . . . . . . . . . . . . sinα新inβ


    . . . . . . .h新in(α + β)
    . . . . = .---------------
    . . . . . . . .sinα新inβ

    i like your part (b), it's much more elegant and shorter than mine. i saw that i could do mine shorter afterwards.

    our part (a)'s are identical, except i called C what you called D, and D what you called C
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