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Trig Problem

  1. Mar 24, 2007 #1
    Hey guys, got a small problem and need some help :frown:

    The problem statement, all variables and given/known data

    Show that

    [tex] \arctan{\left( \frac{x}{2} \right)} = \arccos{\left(\frac{2}{\sqrt{4+x^2}}\right)} \ \mbox{for x}\epsilon\mbox{R} [/tex]

    The attempt at a solution

    Honestly Im pretty stumped from the very beginning....

    The only thing I can currently think of to do is go...

    [tex] \arctan{\frac{x}{2}} = \frac{\arcsin{\frac{x}{2}}}{\arccos{\frac{x}{2}}} [/tex]

    but Im not sure if that is even correct....

    Even still, if that is valid, Im still pretty unsure what Im meant to do next..

    Any hints to point me in the right direction would be much appreciated :redface:

    I hope I did the Latex stuff right, its my first time using it..
    Last edited: Mar 25, 2007
  2. jcsd
  3. Mar 24, 2007 #2


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    Draw a right-triangle and label the sides until you can form a triangle which give s the relationship that you are looking for in your equation. This may give you another formulable relationship which permits you to solve the problem.
  4. Mar 24, 2007 #3


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    I misunderstood the meaning of the problem. You are probably looking for identity relationships to PROVE that your given relation is an identity. Of course, when you draw a right-triangle, you will be able to derive the relationship but you are trying to use a trail of identities to prove this. I wish I could offer better help.

    The best that I could do right now is to draw a triangle; I label one of the non-right angles; the side opposite I give as "x"; the side between the referenced angle and the right-angle I give as length 2; pythagorean theorem gives the hypotenuse as (4 + x^2)^(1/2). Continued reference to this triangle gives the arcos expression which you wanted -------- I am not well with being able to prove as you wanted, but maybe you might be able to now?
  5. Mar 24, 2007 #4
    Ok, taking what you said, here is what I did...


    [tex] \arctan{\left(\frac{x}{2}\right)} = \arccos{\left(\frac{2}{\sqrt{4+x^2}}\right)} [/tex]

    [tex] \cos{\left( \arctan{\left(\frac{x}{2}\right)}\right)} = \frac{2}{\sqrt{4+x^2}} [/tex]

    So then I end up with this triangle....


    \arctan{\left(\frac{x}{2}\right)} = \arccos{\left(\frac{2}{\sqrt{2+x^2}}\right)}[/tex]

    \frac{x}{2}\ =\ \tan{\left(\arccos{\left(\frac{2}{\sqrt{4+x^2}}\right)}\right)}[/tex]

    So then I end up with this triangle....

    This does seem to show that the initial statement is true, however it is quite different to any other 'show' question I have done before

    EDIT: Sorry, Latex is being mean, Im trying to fix it up to show what I really want....
    EDIT2: Yay, I think I've got it all as to how I want it. Wish I had a scanner that worked, woulda made this much easier :P
    Last edited: Mar 25, 2007
  6. Mar 25, 2007 #5


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    Are you sure you've copies the problem correctly?
    What if [tex]x = -2[/tex]?
    [tex]\arctan \left( \frac{-2}{2} \right) = \arctan (-1) = -\frac{\pi}{4}[/tex]
    [tex]\arccos \left( \frac{2}{\sqrt{4 + (-2) ^ 2}} \right) = \arccos \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4}[/tex]
    [tex]\arctan \left( \frac{-2}{2} \right) \neq \arccos \left( \frac{2}{\sqrt{4 + (-2) ^ 2}} \right)[/tex] (Q.E.D)
  7. Mar 25, 2007 #6
    Yup, I definately copied the problem down correctly... weird huh :(
  8. Mar 25, 2007 #7
    Either you are not working in principle values or the question is copied down incorrectly.
    because [itex] cos \frac{\pi}{4}= cos \frac{- \pi}{4} [/itex]

    but the inverse doesnt hold as [itex]cos^{-1} \mbox{has principle range as} [0,\pi] [/itex]
  9. Mar 25, 2007 #8
    Show that [tex] \arctan{\left( \frac{x}{2} \right)} = \arccos{\left(\frac{2}{\sqrt{4+x^2}}\right)} \ \mbox{for x}\epsilon\mbox{R} [/tex]

    ..is the question, character for character :frown:
  10. Mar 25, 2007 #9


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    Well, then, the problem cannot be proven. Because, it's... you know, false. o:)
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