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Trig Problem

  • #1
Hey guys, got a small problem and need some help :frown:


Homework Statement

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:

Answers and Replies

  • #2
symbolipoint
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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.
 
  • #3
symbolipoint
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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?
 
  • #4
VietDao29
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:confused:
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]
Whereas:
[tex]\arccos \left( \frac{2}{\sqrt{4 + (-2) ^ 2}} \right) = \arccos \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4}[/tex]
So:
[tex]\arctan \left( \frac{-2}{2} \right) \neq \arccos \left( \frac{2}{\sqrt{4 + (-2) ^ 2}} \right)[/tex] (Q.E.D)
:smile:
 
  • #5
Yup, I definately copied the problem down correctly... weird huh :(
 
  • #6
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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]
 
  • #7
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:
 
  • #8
VietDao29
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Well, then, the problem cannot be proven. Because, it's... you know, false. o:)
 

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