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Math Help - Square root promblem

  1. #1
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    Square root promblem

    Err, my algera is bad. Im looking over a line integral promblem, and i understand everything, even the calculus, but im stuck on the algebra. here is the promblem.

     \sqrt{((-4sint^2)+(4cost^2))}dt = 4dt

    I understand that  sint^2\ and\ cost^2 = 1 but thought the answer should be  sqrt(32) . Any step by step explanations would be appreciated, thanks.
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  2. #2
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    More information please

    Hello purplerain
    Quote Originally Posted by purplerain View Post
    Err, my algera is bad. Im looking over a line integral promblem, and i understand everything, even the calculus, but im stuck on the algebra. here is the promblem.

     \sqrt{((-4sint^2)+(4cost^2))}dt = 4dt

    I understand that  sint^2\ and\ cost^2 = 1 but thought the answer should be  sqrt(32) . Any step by step explanations would be appreciated, thanks.
    You will have to set the question out more carefully if you want help here. I assume that this is an integration question (in which case it should be in the Calculus Form), but your use of notation is very unclear.

    I can't see how you get 4dt. It's \sin^2t + \cos^2t = 1, not -\sin^2t+\cos^2t. And in any event, when you take the square root, you'll get 2, not 4.

    The only way you can get 4dt is if the expression is \sqrt{(-4\sin t)^2+(4\cos t)^2}\;dt. Is that what you mean?

    The answer you quote, \sqrt{32}, suggests it may be a definite integral. Can we see the original question, please?

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello purplerainYou will have to set the question out more carefully if you want help here. I assume that this is an integration question (in which case it should be in the Calculus Form), but your use of notation is very unclear.

    I can't see how you get 4dt. It's \sin^2t + \cos^2t = 1, not -\sin^2t+\cos^2t. And in any event, when you take the square root, you'll get 2, not 4.

    The only way you can get 4dt is if the expression is \sqrt{(-4\sin t)^2+(4\cos t)^2}\;dt. Is that what you mean?

    The answer you quote, \sqrt{32}, suggests it may be a definite integral. Can we see the original question, please?

    Grandad
    Hi, yes, i it was an integration question, but as you know, a lot of calculus is algebra. I just had a question about the algebra, but i will post the full question in the appropriate forum.
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  4. #4
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    Quote Originally Posted by purplerain View Post
    Err, my algera is bad. Im looking over a line integral promblem, and i understand everything, even the calculus, but im stuck on the algebra. here is the promblem.

     \sqrt{((-4sint^2)+(4cost^2))}dt = 4dt

    I understand that  sint^2\ and\ cost^2 = 1 but thought the answer should be  sqrt(32) . Any step by step explanations would be appreciated, thanks.
    Be aware that you may be expected to simplify your answer as \sqrt{32} is not fully simplified \sqrt{32} = 4\sqrt{2}

    Because of your identity I will assume you mean sin^2(t) not sin(t^2)

    4cos^2t-4sin^2t = 4(cos^2t-sin^2t)

    cos^2t-sin^2t = cos(2t)

    cos(2t)

    \sqrt{4} \sqrt{cos(2t)} = 2\sqrt{cos(2t)}
    Last edited by e^(i*pi); November 21st 2009 at 09:11 AM. Reason: see post below
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  5. #5
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    Hello everyone
    Quote Originally Posted by e^(i*pi) View Post
    ...

    cos^2t-sin^2t = cos^2(2t)...
    In case you need to use this expression, note that \cos^2t - \sin^2t = \cos2t, not \cos^22t.

    Grandad
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  6. #6
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    Quote Originally Posted by Grandad View Post
    Hello everyoneIn case you need to use this expression, note that \cos^2t - \sin^2t = \cos2t, not \cos^22t.

    Grandad
    How could I not notice that

    *edited above*
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