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    (Original post by steve2005)
    My answer to the OP is that the negative disappears because of the modulus sign, and that is the only reason.
    Yes. The OP said that, and nuodai agreed with him, and then went on to say something further.
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    Maths wars. Who is correct? find out next week.
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    Looking at nuodai post again I can see that I was wrong to criticise .

    I misread his post in that i did not take account of "any more"

    I agree that the modulus makes the expression positive and that since both numerator and denominator are squares then brackets can be used rather than the modulus sign.




    Edit: I was wrong to bow to pressure , the blue brackets should be replaced by the modulus sign. Note posts 24 and 25.
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    (Original post by steve2005)
    Right, \displaystyle \dfrac{1}{2} \ln \left| \dfrac{(1 + \sin x)^2}{-\cos^2 x} \right| = \dfrac{1}{2} \ln \left| \dfrac{(1 + \sin x)^2}{\cos^2 x} \right|, I'll agree with you there. And also \displaystyle \dfrac{1}{2} \ln \left| \dfrac{(1 + \sin x)^2}{\cos^2 x} \right| = \dfrac{1}{2} \ln \left[ \dfrac{(1 + \sin x)^2}{\cos^2 x} \right], because what's in the modulus is always positive. However, it's not true that \displaystyle \dfrac{1}{2} \ln \left[ \dfrac{(1 + \sin x)^2}{\cos^2 x} \right] = \ln \left[ \dfrac{1 + \sin x}{\cos x} \right]; take x = \pi as an example... this is because you have to take the positive square root, so you have to put a modulus back in again, giving you \displaystyle \ln \left| \dfrac{1 + \sin x}{\cos x} \right|.

    I didn't follow the rest of the thread, so if I'm not addressing anything here let me know.
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    (Original post by nuodai)
    Right, \displaystyle \dfrac{1}{2} \ln \left| \dfrac{(1 + \sin x)^2}{-\cos^2 x} \right| = \dfrac{1}{2} \ln \left| \dfrac{(1 + \sin x)^2}{\cos^2 x} \right|, I'll agree with you there. And also \displaystyle \dfrac{1}{2} \ln \left| \dfrac{(1 + \sin x)^2}{\cos^2 x} \right| = \dfrac{1}{2} \ln \left[ \dfrac{(1 + \sin x)^2}{\cos^2 x} \right], because what's in the modulus is always positive. However, it's not true that \displaystyle \dfrac{1}{2} \ln \left[ \dfrac{(1 + \sin x)^2}{\cos^2 x} \right] = \ln \left[ \dfrac{1 + \sin x}{\cos x} \right]; take x = \pi as an example... this is because you have to take the positive square root, so you have to put a modulus back in again, giving you \displaystyle \ln \left| \dfrac{1 + \sin x}{\cos x} \right|.

    I didn't follow the rest of the thread, so if I'm not addressing anything here let me know.

    I agree with you 100%

    Two sentences deleted because they contained wrong material.
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    (Original post by steve2005)
    The modulus sign should be used throughout the entire problem and hence the use of square brackets is wrong.
    There's no reason why the modulus sign should be used when what's within it is always positive. Like, if x \in \mathbb{R}, there'd be no point writing \ln \left| x^2 \right| when you can write \ln x^2, but you would need to write 2\ln |x| if you were going to take the power of 2 outside the log. It doesn't really matter though; I mean the modulus sign does no harm, but it needn't be there for the sake of it :p: Where have you got this idea from?
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    (Original post by nuodai)
    There's no reason why the modulus sign should be used when what's within it is always positive. Like, if x \in \mathbb{R}, there'd be no point writing \ln \left| x^2 \right| when you can write \ln x^2, but you would need to write 2\ln |x| if you were going to take the power of 2 outside the log. It doesn't really matter though; I mean the modulus sign does no harm, but it needn't be there for the sake of it :p: Where have you got this idea from?
    I agree.
 
 
 
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