Help Needed: Solving Trig Equation Involving Arcsin & Arccos

[][nstigator]

Hey guys, got a small problem and need some help Homework Statement

Show that

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

The attempt at a solution

Honestly I am pretty stumped from the very beginning...

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

$$\arctan{\frac{x}{2}} = \frac{\arcsin{\frac{x}{2}}}{\arccos{\frac{x}{2}}}$$

but I am not sure if that is even correct...

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

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

I hope I did the Latex stuff right, its my first time using it..

 

[symbolipoint]

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.

 

[symbolipoint]

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?

 

[VietDao29]

Are you sure you've copies the problem correctly?
What if $$x = -2$$?
$$\arctan \left( \frac{-2}{2} \right) = \arctan (-1) = -\frac{\pi}{4}$$
Whereas:
$$\arccos \left( \frac{2}{\sqrt{4 + (-2) ^ 2}} \right) = \arccos \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4}$$
So:
$$\arctan \left( \frac{-2}{2} \right) \neq \arccos \left( \frac{2}{\sqrt{4 + (-2) ^ 2}} \right)$$ (Q.E.D)

 

[][nstigator]

Yup, I definitely copied the problem down correctly... weird huh :(

 

[f(x)]

Either you are not working in principle values or the question is copied down incorrectly.
because $cos \frac{\pi}{4}= cos \frac{- \pi}{4}$

but the inverse doesn't hold as $cos^{-1} \mbox{has principle range as} [0,\pi]$

 

[][nstigator]

Show that $$\arctan{\left( \frac{x}{2} \right)} = \arccos{\left(\frac{2}{\sqrt{4+x^2}}\right)} \ \mbox{for x}\epsilon\mbox{R}$$

..is the question, character for character

 

[VietDao29]

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