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Prove arccos(-x)+arccos(x) = pi

  1. Mar 20, 2015 #1

    GreyNoise

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    1. The problem statement, all variables and given/known data
    From an old edition of the Anton calculus text, I am asked to prove
    cos-1(-x) + cos-1(x) = π
    or equivalently
    cos-1(-x) = π - cos-1(x)

    2. Relevant equations


    3. The attempt at a solution
    Earlier I proved that sin-1(x) was odd by noting sin-1(sin(x)) = -sin-1(sin(-x)), so I have tried to start the problem from the same approach, but none of my attempts have been insightful.
    I started with cos-1(cos(x)) = x and cos-1(cos(-x)) = -x but I am not going anywhere with it. I'm stuck. It makes geometric sense that the sum cos-1(-x) + cos-1(x) is in fact π when viewing a graph of cos-1(x). But how do I prove this one? It appears that I am not supposed to use derivatives of the inverses to solve this (they are covered after this section in the text). It is from Anton's Brief Edition 1981, p 462. Any hints or the like are appreciated.
     
  2. jcsd
  3. Mar 20, 2015 #2

    epenguin

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    :blushing:
     
    Last edited: Mar 20, 2015
  4. Mar 20, 2015 #3

    LCKurtz

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    Try calling ##\alpha = \cos^{-1}(-x)## and ##\beta = \cos^{-1}(x)##. Convince yourself that the angles on both sides of your equation are in ##[0,\pi]## and then check that they have equal cosines.
     
  5. Mar 20, 2015 #4

    GreyNoise

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    Thank you epenguin and LCKurtz. I thought over your hints, but I was unable to wrap my head around them; I tried the following.

    (1) [itex] \alpha [/itex] and [itex] \beta [/itex] are angles of a right triangle : given

    (2) [itex] \alpha + \beta + \frac{\pi}{2} = \pi [/itex] : sum of angles is [itex] \pi [/itex] for a triangle

    (3) [itex] \cos^{-1}(\cos(\alpha)) + \cos^{-1}(\cos(\beta)) + \frac{\pi}{2} = \pi [/itex] : substitute definition for inverse

    (4) [itex] \cos^{-1}(\cos(\alpha)) + \cos^{-1}(\cos(\pi/2-\alpha)) + \frac{\pi}{2} = \pi [/itex] : because [itex] \beta = \frac{\pi}{2} - \alpha [/itex]

    (5) [itex] \cos^{-1}(\cos(\alpha)) + \cos^{-1}(\sin(\alpha)) + \frac{\pi}{2} = \pi [/itex] : because [itex] \cos(\pi/2-\alpha) = \sin(\alpha) [/itex]

    From step (3), I also tried
    (4) [itex] \cos^{-1}(\cos(\alpha)) + \cos^{-1}(\cos(-\beta)) + \frac{\pi}{2} = \pi [/itex] : because [itex] \cos(x) [/itex] is even

    But setting

    [itex] -\cos(\alpha) = \cos(-\beta)) [/itex]

    seems to beg the question and I don't know how get through that. Can you tell me what I am missing here?
     
  6. Mar 20, 2015 #5

    Ray Vickson

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    For notational convenience below, say you want to prove ##\arccos(a) + \arccos(-a) = \pi##. Draw a picture: locate the points ##P_1, P_2## on the unit circle whose x-coordinates are ##x= a## and ##x = -a##. Now stare at the picture.
     
  7. Mar 20, 2015 #6

    LCKurtz

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    @GreyNoise: You are trying to show ##\alpha = \pi - \beta##. Take the cosine of both sides, written just like that and simplify it.
     
  8. Mar 20, 2015 #7

    GreyNoise

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    I think I got it! The unit circle will carry x=a to x=-a if we sweep [itex] \pi [/itex] rads (thnx Ray V.), so the angles [itex] \cos^{-1}(x) [/itex] and [itex] \cos^{-1}(-x) [/itex] are supplementary (thnx LCKurtz). So I proceeded as follows:

    (1) [itex] \alpha + \beta = \pi [/itex]
    (2) [itex] \cos^{-1}(\cos\alpha) + \cos^{-1}(\cos\beta) = \pi [/itex] : definition of inverse
    (3) [itex] \beta = \pi - \alpha [/itex] : subtract [itex] \alpha [/itex] from (1)
    (4) [itex] \cos(\pi - \alpha) = \cos\pi\cos\alpha+\sin\pi\sin\alpha [/itex] : trig identity
    (5) [itex] \cos(\pi - \alpha) = -\cos\alpha [/itex] : simplify (4)
    (6) [itex] \cos\beta = -\cos\alpha [/itex] : sub (3) into (5)
    (7) [itex] \cos^{-1}(\cos\alpha) + \cos^{-1}(-\cos\alpha) = \pi [/itex] : sub (6) into (2)
    (8) [itex] x = \cos\alpha [/itex] so
    (9) [itex] \cos^{-1}(x) + \cos^{-1}(-x) = \pi [/itex]

    Wella
     
  9. Mar 20, 2015 #8

    LCKurtz

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    That's what you are trying to prove. You don't usually start with that, unless you are claiming it is true from the geometry. But in that case you don't need the stuff below.

    So you ended up with what you started with.

    So, one more time, try taking the cosine both sides of ##\alpha =\pi - \beta##. If the angles have equal cosines in that range then the angles are equal. It is two simple steps and you don't have to use any geometry.
     
  10. Mar 21, 2015 #9

    GreyNoise

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    Is this what you have in mind for me LCK?

    [itex] \alpha = \pi - \beta [/itex]
    [itex] \cos\alpha = cos(\pi - \beta) [/itex]
    [itex] \cos\alpha = -cos\beta\hspace{10mm}[/itex]:by difference trig identity so
    [itex] \cos^{-1}(\cos\alpha) = \cos^{-1}(-cos\beta) [/itex]

    Now sub back for [itex] \alpha [/itex] and [itex] \beta [/itex] at the top I get

    [itex] \cos^{-1}(-cos\beta) = \pi - \cos^{-1}(cos\beta) [/itex]

    which has the form

    [itex] \cos^{-1}(-x) = \pi - \cos^{-1}(x) [/itex]
     
  11. Mar 21, 2015 #10
    What about differentiating arcos(-x) and arcos(x) separately and noting that the derivatives are the equal?
     
  12. Mar 21, 2015 #11

    LCKurtz

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    No. You are still overcomplicating it and starting with what you are trying to prove. We are starting by calling$$
    \cos^{-1}(-x) = \alpha,~\cos^{-1}(x) = \beta$$
    You are trying to show ##\alpha = \pi - \beta## by showing they have equal cosines. So you start by taking the cosines of both sides. The left side gives you ##\cos\alpha## and the right side gives you ##\cos(\pi - \beta) = -\cos(\beta)##, as you have shown. So far so good. So now if ##\cos\alpha = -\cos(\beta)## you are done. From your given, what is ##\cos\alpha## and what is ##\cos\beta##? Give the answers in terms of ##x##. Does ##\cos\alpha=-\cos\beta##?
     
  13. Mar 21, 2015 #12

    GreyNoise

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    This section calls for to not use differentiation; I've done it that way before (long time ago), but this is a new restriction, no derivatives.
     
  14. Mar 21, 2015 #13

    GreyNoise

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    Yes, I did get the sinking feeling I was begging the question after I posted
    So how about this?

    given:[itex]\hspace{10mm}\cos^{-1}(-x) = \alpha[/itex] and [itex]\cos^{-1}(x) = \beta[/itex]
    show:[itex]\hspace{10mm}\alpha = \pi - \beta [/itex]

    from given
    [itex]\cos\alpha = -x[/itex]
    [itex]\cos\beta = x\hspace{5mm}[/itex] so
    [itex]\cos\alpha = -cos\beta[/itex]

    [itex]\cos\alpha = \cos(\pi - \beta) = -cos\beta\hspace{10mm}[/itex]:trig diff id
    [itex]\cos\alpha = -cos\beta\hspace{10mm}[/itex]:same as given

    so
    [itex] \alpha = \pi - \beta [/itex]
     
  15. Mar 21, 2015 #14

    LCKurtz

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    Yes, you have now got all the steps, but that is a very awkward arrangement of steps for an argument. Since you seem to now have it figured out, I will suggest a more reasonable way to write it up:

    Let ##\alpha=\cos^{-1}(-x)## and ##\beta = \cos^{-1}x##. To show ##\alpha = \pi -\beta##.

    ##\alpha## and ##\pi - \beta## are both in ##[0,\pi]##, so if their cosines are equal, the angles are equal.

    ##\cos\alpha = -x##
    ##cos(\pi - \beta) = \cos\pi\cos\beta+\sin\pi\sin\beta = -\cos\beta = -x##
    Therefore ##\cos\alpha = \cos(\pi -\beta)## so ##\alpha = \pi - \beta##, and the proof is finished
     
  16. Mar 21, 2015 #15

    GreyNoise

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    Got it LCK, thnx!
     
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