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Math Help - Trigonometric Equations - Different Methods/Different Results

  1. #1
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    Trigonometric Equations - Different Methods/Different Results

    This is the equation:

    1 + Sin(theta) = 2 Cos squared(theta)

    The method I used goes like this:

    1 + Sin(theta) = 2[1 - Sin squared(theta)] uses Pythagorean Identities


    1 + Sin(theta)
    ------------------------------ = 2 divide by [1 - Sin squared(theta)]
    (1 + Sin(theta)) (1 - Sin(theta))


    1
    --------------- = 2
    (1 - Sin(theta))


    1 - Sin(theta) = 1/2

    -Sin(theta) = -1/2

    Sin(theta) = 1/2

    theta = PI/6 + 2k(PI) or theta = 5PI/6 + 2k(PI)

    restrict the solution to 0 < theta < 2PI

    therefore the solution is {PI/6, 5PI/6}




    HOWEVER you get a MORE complete solution if you work via this method:

    1 + Sin(theta) = 2 Cos squared(theta) original equation

    1 + Sin(theta) = 2[1 - Sin squared(theta)] uses Pythagorean Identities

    1 + Sin(theta) = 2 - 2 Sin squared(theta) multiplies out the right side

    2 Sin squared(theta) + Sin(theta) - 1 = 0 rearranges the equation

    (2 Sin(theta) - 1) (Sin(theta) + 1) = 0 factors left side


    2 Sin(theta) - 1 = 0 or Sin(theta) + 1 = 0

    Sin(theta) = 1/2 or Sin(theta) = -1

    theta = {PI/6, 5PI/6} or theta = 3PI/2

    Clearly the second method yields a more complete answer, and I verified it with a TI-89. Yet the first method is algebraicly sound, as far as I can tell. Any thoughts on this?
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  2. #2
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    Because you cannot divide by,
    1-\sin^2 \theta!!!!

    Because it might be zero!
    Then you divided by zero!
    (Oh no!)

    Once you divide by zero you get strange results.
    That was you fault.
    -----
    Instead:

    1+\sin x=2(1-\sin^2 x)
    Thus,
    (1+\sin x)-2(1+\sin x)(1-\sin x)=0
    Factor,
    (1+\sin x)[1-2(1-\sin x)]=0
    Thus,
    (1+\sin x)(2\sin x-1)=0
    Now set each factor equal to zero.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by spiritualfields
    This is the equation:

    1 + Sin(theta) = 2 Cos squared(theta)

    The method I used goes like this:

    1 + Sin(theta) = 2[1 - Sin squared(theta)] uses Pythagorean Identities


    1 + Sin(theta)
    ------------------------------ = 2 divide by [1 - Sin squared(theta)]
    (1 + Sin(theta)) (1 - Sin(theta))
    This is only valid when both 1 + \sin(\theta) \ne 0 and 1 - \sin(\theta) \ne 0, so these cases need to be checked separately.

    RonL
    Last edited by CaptainBlack; August 2nd 2006 at 01:12 PM.
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  4. #4
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    Thank you, The Perfect Hacker and Captain Black. Interestingly, with this particular equation, the value that did not show up when I divided by (1 + Sin(theta)) (1 - Sin(theta)) was 3PI/2, which in fact is the value that creates the divide by zero situation for this equation ... (1 + Sin(3PI/2)).

    Ed
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