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Math Help - Simplification of Trig Functions TWO

  1. #1
    Member classicstrings's Avatar
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    Simplification of Trig Functions TWO

    Hi, I have

    [(sin^2theta) - (cos^2theta) ]/ (sin(theta) + cos(theta))

    And I need to simplify it to sin(theta) - cos(theta).

    How do I do that? Thanks!
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by classicstrings
    Hi, I have

    [(sin^2theta) - (cos^2theta) ]/ (sin(theta) + cos(theta))

    And I need to simplify it to sin(theta) - cos(theta).

    How do I do that? Thanks!
    simply divide
    for siplification, use sin(theta)=a;cos(theta)=b
    you could divide only if a in not equal to b

    Keep Smiling
    Malay
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  3. #3
    Member classicstrings's Avatar
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    So,

    (a^2 - b^2)/ (a+b)

    Right?
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by classicstrings
    So,

    (a^2 - b^2)/ (a+b)

    Right?
    yes
    you needn't use any trigonometric formulae here. It is simple algebraic division

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  5. #5
    Member classicstrings's Avatar
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    Done thanks!
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  6. #6
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    You can also factor, instead of long divsion.
    \frac{(\sin x+\cos x)(\sin x-\cos x)}{\sin x+\cos x}

    Quote Originally Posted by malaygoel
    you could divide only if a in not equal to b
    You mean a\not = -b
    Last edited by ThePerfectHacker; September 8th 2006 at 08:44 AM.
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  7. #7
    Member classicstrings's Avatar
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    Ah yes factoring, thanks for the 2nd opinion!
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by classicstrings
    Ah yes factoring, thanks for the 2nd opinion!
    The interesting thing here is that Malay would not have suggested that
    you simply divide if he had not already recognised that the numerator
    factored from the difference of two squares form that it has and that
    one of these factors was the denominator.

    (At least that is my take on what was going on)

    RonL
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