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Thread: can u solve this in less than 8 steps? double angle trig Identities

  1. #1

    Thread Starter
    Registered User struntz's Avatar
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    can u solve this in less than 8 steps? double angle trig Identities

    hello everyone!
    i solved this trig identity but it took 8 steps! and i was thinking htere had to be an easier way and would like to find out...
    here is what i did...
    sec2x = sec^2x + sec^4x
    ----------------------
    2 + sec^2x - sec%^4x

    = sec^2x(1 + sec^2x)
    -----------------------------
    (2-sec^2x)(1+sec^2x)

    = sec^2x
    -----------
    2-sec^2x

    = 1/cos^2x
    ------------
    2-1/cos^2x

    = 1/cos^2x
    ---------------------------
    2cos^2x - 1
    -----------------
    cos^2x

    (1/cos2x)(cos^2x/2cos^2x -1)

    = 1
    -----------
    2 cos^2x-1 //double angle, cos2x = 2cos^2x-1

    = 1
    --------
    cos2x

    = sec2x

    if any of u can do it a differnet and faster way post it, thanks~!

  2. #2
    Hyperactive Member thinktank2's Avatar
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    Re: can u solve this in less than 8 steps? double angle trig Identities

    (Sec[x])^2 + (Sec[x])^4
    ----------------------------
    2 + (Sec[x])^2 - (Sec[x])^4


    //--- Step 1 ---

    = (Sec[x])^2 * [(Sec[x])^2 + 1]
    -----------------------------------
    [(Sec[x])^2 + 1] - [(Sec[x])^4 - 1]


    //--- Step 2 ---
    Dividing Both Numerator and Denominator by [(Sec[x])^2 + 1]

    = [1 + (tan[x])^2] * 1
    -------------------------
    1 - [(Sec[x])^2 - 1]


    //--- Step 3 ---

    = [1 + (tan[x])^2]
    ---------------------
    [1 - (tan[x])^2]


    //--- Step 4 ---
    Multiplying Both Numerator and Denominator by (Cos[x])^2

    = (Cos[x])^2 + (Sin[x])^2
    -----------------------------
    (Cos[x])^2 - (Sin[x])^2


    //--- Step 5 ---

    = 1/Cos[2x]

    //--- Step 6 ---

    = Sec[2x]

  3. #3
    PowerPoster beachbum's Avatar
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    Step 1
    Scroll down

    Step 2
    Copy

    Step 3
    Paste
    Sec[2x]

    done !
    Stuart Laidlaw
    Brightspark Financial Software
    http://www.gstsmartbook.com

  4. #4

  5. #5

    Thread Starter
    Registered User struntz's Avatar
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    hm...

    Think tank, thanks for the responce, but i noticed u solved the proof as if it where

    (Sec[x])^2 + (Sec[x])^4
    ----------------------------
    2 + (Sec[x])^2 - (Sec[x])^4

    but it really is
    sec^2x + sec^4x
    -----------------------
    2 + sec^2x - sec^4x


    or is that also correct becuasde u wrote it with the ( ) 's?

  6. #6
    Addicted Member
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    Well, it depends on what do you consider as a step. I didn't follow your proof but I think a proof with the steps detailed is this:


    //Step 1// multiply by [sec(x)]^-4

    1+[sec(x)]^-2
    ----------------------------------------
    2[sec(x)]^-4 + [sec(x)]^-2 - 1


    //Step 2// convert sec to cos

    1 + [cos(x)]^2
    --------------------------------------------
    2 [[cos(x)]^2]^2 + [cos(x)]^2 - 1


    //Step 3// factor denominator 2[w]^2 + w - 1 == (1+w)(2w-1)

    1
    ---------------------
    2[cos(x)]^2 - 1


    //Step 4// known identity : 2[cos(x)]^2 - 1 = cos(2x)

    1
    --------
    cos(2x)


    Done !

    Actually, copying & pasting is faster, but.....

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