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  1. #1
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    Help Simplifying

    ( sin^3(x) + cos^3(x) ) / ( sin^2(x) - cos^2(x) ) Please help me simplify this one. Giving me a headache.
    Last edited by FaRmBoX; December 14th 2009 at 01:42 PM.
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  2. #2
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    Quote Originally Posted by FaRmBoX View Post
    sin^3(x) + cos^3(x) -------------------- sin^2(x) - cos^2(x) Please help me simplify this one. Giving me a headache.
    What are you trying to ask?

    A. sin^3(x)+cos^3(x)

    Use the sum of two cubes a^3+b^3 = (a+b)(a^2-ab+b^2) and the Pythagorean identity

    B. sin^2(x)-cos^2(x)

    You could use the difference of two squares but it'd get you nowhere really. I would rewrite as -(cos^2(x)-sin^2(x)) and now you can use the double angle identity for cos(2x)
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  3. #3
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    That was my bad, it the first value divided by the 2nd value. I put them one above the other as I typed it and I didnt realize the entire post ended being one line. ( sin^3(x) + cos^3(x) ) / ( sin^2(x) - cos^2(x) )
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by FaRmBoX View Post
    That was my bad, it the first value divided by the 2nd value. I put them one above the other as I typed it and I didnt realize the entire post ended being one line. ( sin^3(x) + cos^3(x) ) / ( sin^2(x) - cos^2(x) )
    \frac{sin^3(x)+cos^3(x)}{sin^2(x)-cos^2(x)}

    Slightly different this time. Again use the sum of two cubes on the top

    sin^3(x)+cos^3(x) = [sin(x)+cos(x)][sin^2(x)-sin(x)cos(x)+cos^2(x)]

    As sin^2(x)+cos^2(x) = 1 we get [sin(x)+cos(x)][1-sin(x)cos(x)]

    As far as I know once sin(x)+cos(x) goes it cannot be reduced further

    This time use the difference of two squares on the bottom to ensure something will cancel in the step after this

    sin^2(x)-cos^2(x) = [sin(x)+cos(x)][sin(x)-cos(x)]

    Bring both parts together

    \frac{[sin(x)+cos(x)][1-cos(x)sin(x)]}{[sin(x)+cos(x)][sin(x)-cos(x)]}
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