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Why is this wrong?

  1. Jan 15, 2008 #1
    Why is this wrong???

    arcos(x)= ∫(-1)/√(1-x) dx = -∫1/√(1-x) dx = -arcsin(x)

    ⇒ arcos(x)=-arcsin(x)

    i know this is wrong, i think its got something to do with constant of integration, could someone pls point out my mistake. thnks
     
  2. jcsd
  3. Jan 15, 2008 #2

    Tom Mattson

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    It's wrong because you forgot the +C at the end of both integrals.

    [tex]\int\frac{-1}{\sqrt{1-x^2}}dx=\arccos(x)+C_1[/tex]

    [tex]-\int\frac{1}{\sqrt{1-x^2}}dx=-\arcsin(x)+C_2[/tex]

    Therefore,

    [tex]\arccos(x)+C_1=-\arcsin(x)+C_2[/tex]

    [tex]\arcsin(x)+\arccos(x)=C_2-C_1[/tex]

    Since [itex]C_1[/itex] doesn't necessarily equal [itex]C_2[/itex], the two inverse trig functions don't necessarily add up to 0.
     
  4. Jan 15, 2008 #3
    but cant c1, c2 be what ever i choose them to be, say 0?
     
  5. Jan 15, 2008 #4

    Tom Mattson

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    No. They are arbitrary constants.
     
  6. Jan 15, 2008 #5
    oh, and also, why is arcsin(x) + arcos(x) = pi/2
     
  7. Jan 15, 2008 #6

    rock.freak667

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    Let A=arcsin(x) and B=arccos(x)

    so that sinA=x and cosB=x

    and the equation is A+B, then consider what sin(A+B) is and check when A and B are both acute.
     
  8. Jan 15, 2008 #7

    Tom Mattson

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    Here's one way to show that. Let [itex]x\in[-1,1][/itex]

    Then:

    [tex]x=x[/tex]

    [tex]x=\cos\left(\frac{\pi}{2}\right)cos(x)+\sin\left(\frac{\pi}{2}\right)\sin\sin^{-1}(x)[/tex]

    [tex]x=\cos\left(\frac{pi}{2}-sin^{-1}(x)\right)[/tex]

    (Here I used the trig identity cos(x-y)=cos(x)cos(y)+sin(x)sin(y))

    Take the inverse cos of both sides, and conclude.
     
  9. Jan 15, 2008 #8

    Gib Z

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    We could have cloncluded that they only differ by a constant from the integrals we just saw: The functions can only differ by a constant. Let x=0, since the function (arcsin x - arccos x) is a constant, it takes that value for all points.
     
  10. Jan 16, 2008 #9

    I don't know gib z, the constants are arbitrary in order to integrate since the derivative of a constant is zero

    if we take sinx - cosx for several values for x, like the most simple 0 and 1, sinx - cosx will not be constant
     
  11. Jan 16, 2008 #10

    Tom Mattson

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    But it's not sin(x)-cos(x) we're interested in. It's arcsin(x)-arccos(x).
     
  12. Jan 16, 2008 #11
    since arc is the inverse function I tought that arcsin(x) +- arccos(x) = constant for several values of x <==> sin(x) +- cos(x) = variable for several values of x ==> absurdum

    so we can use the trig. functions in our analisys, what do you think Tom?
     
    Last edited: Jan 16, 2008
  13. Jan 16, 2008 #12

    Gib Z

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    So you don't like my proof >.<" ?
     
  14. Jan 16, 2008 #13
    I dont have nothing against :smile:, I just think that it is wrong :shy:, but I may be wrong, let's wait for tom mattson opinion
     
  15. Jan 16, 2008 #14

    Gib Z

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  16. Jan 16, 2008 #15

    Mute

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    That is incorrect. arcsin(x) +/- arccos(x) = constant in no way implies sin(x) +/- cos(x) is a constant. I'm not sure, but it looks like in your first sentence you interpret "arc" to be some sort of operator such that

    arc(cos(x)) := arccos(x)
    arc(sin(x)) := arcsin(x)

    and then you conclude that arc(cos(x) +/- sin(x)) = arccos(x) +\- arcsin(x), which isn't the case at all. arcsine and arccosine are simply functions with the property such that

    arccos(cos(x)) = x
    cos(arccos(x)) = x

    and similarly for arcsine. (Note that the above first expression is really only 100% true in the first quandrant of the cartesian plane. There will be additional constants for the other quadrants.)

    As GibZ stated, the integral calculations above clearly show arcsin(x) and arccos(x) differ only by a constant. Choose any value of x and evaluate the expression - it will be the same no matter what value of x you choose.
     
  17. Jan 16, 2008 #16
    I meant: if you are considering that arcsin(x) +- arccos(x) = constant for several values of x and we know that sin(x) +- cos(x) = variable for several values of x, I think both situations imply an absurd

    arcsin(1/2) = 30, arccos(1/2) = 60 ==> arcsin(0.5) - arccos(0.5) = -30

    arcsin(sqrt(2)/2) = 45, arccos(sqrt(2)/2) = 45 ==> arcsin(sqrt(2)/2) - arccos(sqrt(2)/2) = 0

    so the constant C is not allways the same if you input the same value of x in both functions for all cases of x

    why this conclusion is wrong? I think I gave a concrete counter-proof, or not?
     
    Last edited: Jan 16, 2008
  18. Jan 16, 2008 #17

    this is exactly the conclusion I was trying to show, but my english is not so good :smile:
     
  19. Jan 16, 2008 #18

    Mute

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    The expression in question is arccos(x) + arcsin(x), which is pi/2 always (within the domain of definition of arcos and arcsin). arccos(x) - arcsin(x) need not be (and, indeed is not) a constant.

    Think about it geometrically, if you like:

    For a right-angled triangle with hypotenuse C, base A and height B, let x be the angle between the base and the hypotenuse and y be the angle between the vertical piece of the hypotenuse. By geometry, x + y = pi/2 radians (90 degrees).

    Now, arcsin(A/C) = x, and arccos(A/C) = y. Hence, arcsin(A/C) + arccos(A/C) = x + y = pi/2.
     
  20. Jan 16, 2008 #19

    Gib Z

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    Instead of taking the integral approach, I'll do the differentiation proof.

    Let [itex]f(x) = \arcsin x + \arccos x[/itex].

    Then,
    [tex]\frac{df(x)}{x} = \frac{1}{\sqrt{1-x^2}} + \frac{-1}{\sqrt{1-x^2}} = 0[/tex].

    Since the derivative is 0 everywhere, the value of f(x) never changes. Letting x= [itex]pi/2[/itex], [itex]f(x) = \pi /2[/itex]. Q.E.D.
     
  21. Jan 17, 2008 #20

    Read carefully and you'll see that I was talking about this GIBz statement:


    This is the expression in question in my messages

    But you're right in one point: I've tried to generalize to +-, and this is a mistake.
     
    Last edited: Jan 17, 2008
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