MHB Understanding the restriction on x in arccos(x) = arctan(x)

[karush]

$\arccos\left({x}\right)=\arctan\left({x}\right)$

Explain why $x$ cannot negative or $0$

I assume that

$$\cos\left({\theta}\right)=\tan\left({\theta}\right)$$

This intersects in Q1 but why never negative or 0?

 

[I like Serena]

$\arccos\left({x}\right)=\arctan\left({x}\right)$

Explain why $x$ cannot negative or $0$

I assume that

$$\cos\left({\theta}\right)=\tan\left({\theta}\right)$$

This intersects in Q1 but why never negative or 0?

Hi karush! (Smile)

I'm afraid that $\cos\left({\theta}\right)=\tan\left({\theta}\right)$ is something entirely different.
I suggest to make a drawing of both $\arccos x$ and $\arctan x$ in the same graph.
It should show that their intersection can only be in Q1.

 

[karush]

So you mean simply because the graph intersects in Q1 that explains why? .
They did show a graph of that with the problem

 

[MarkFL]

As the inverse trigonometric functions are normally defined, where do the domains of the two given functions intersect? As a follow-up, can you then find the exact value of $x$?

 

[suluclac]

I believe that the arccosine covers both I & II quadrants. However, the arctangent covers both I & III quadrants.
Clearly, arccosine doesn't intersect the origin (arctangent on the other hand, can).
Therefore, in the I quadrant, they clearly intersect.

 

[Greg]

The range of $\arctan(x)$ is between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ (inclusive), i.e. quadrants I and IV.

 

[suluclac]

greg1313, I thought arctangent is in quadrants I and III.

 

[MarkFL]

greg1313, I thought arctangent is in quadrants I and III.

Quadrants I and III are where the tangent function is positive...whereas the standard chosen domain for the tangent function in which it is one-to-one and the concept of an inverse is meaningful is:

$$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$

 

[suluclac]

Yes, I understand that the tangent is positive on quadrants I & III, but regardless of the arctangent going through the origin, the lines extend to quadrants I & III. Perhaps I'm not understanding the point here.