Trig Problem

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.

 

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?

 

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)

 

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]

 

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