Editing Inverse trigonometric functions (section)

====Detailed example and explanation of the "plus or minus" symbol {{math|±}} ====

The solutions to <math>\cos \theta = x</math> and <math>\sec \theta = x</math> involve the "plus or minus" symbol <math>\,\pm,\,</math> whose meaning is now clarified. Only the solution to <math>\cos \theta = x</math> will be discussed since the discussion for <math>\sec \theta = x</math> is the same. 
We are given <math>x</math> between <math>-1 \leq x \leq 1</math> and we know that there is an angle <math>\theta</math> in some interval that satisfies <math>\cos \theta = x.</math> We want to find this <math>\theta.</math> The table above indicates that the solution is 
<math display="block">\,\theta = \pm \arccos x+2 \pi k\, \quad \text{ for some }k \in \Z</math>
which is a shorthand way of saying that (at least) one of the following statement is true: <br />
#<math>\,\theta = \arccos x+2 \pi k\,</math> for some integer <math>k,</math> <br/>or 
#<math>\,\theta =-\arccos x+2 \pi k\,</math> for some integer <math>k.</math> 
As mentioned above, if <math>\,\arccos x = \pi\,</math> (which by definition only happens when <math>x = \cos \pi = -1</math>) then both statements (1) and (2) hold, although with different values for the integer <math>k</math>: if <math>K</math> is the integer from statement (1), meaning that <math>\theta = \pi+2 \pi K</math> holds, then the integer <math>k</math> for statement (2) is <math>K+1</math> (because <math>\theta = -\pi+2 \pi (1+K)</math>). 
However, if <math>x \neq -1</math> then the integer <math>k</math> is unique and completely determined by <math>\theta.</math> 
If <math>\,\arccos x = 0\,</math>  (which by definition only happens when <math>x = \cos 0 = 1</math>) then <math>\,\pm\arccos x = 0\,</math> (because <math>\,+ \arccos x = +0 = 0\,</math> and <math>\,-\arccos x = -0 = 0\,</math> so in both cases <math>\,\pm \arccos x\,</math> is equal to <math>0</math>) and so the statements (1) and (2) happen to be identical in this particular case (and so both hold). 
Having considered the cases <math>\,\arccos x = 0\,</math> and <math>\,\arccos x = \pi,\,</math> we now focus on the case where <math>\,\arccos x \neq 0\,</math> and <math>\,\arccos x \neq \pi,\,</math> So assume this from now on. The solution to <math>\cos \theta = x</math> is still
<math display="block">\,\theta = \pm \arccos x+2 \pi k\, \quad \text{ for some }k \in \Z</math>
which as before is shorthand for saying that one of statements (1) and (2) is true. However this time, because <math>\,\arccos x \neq 0\,</math> and <math>\,0 < \arccos x < \pi,\,</math> statements (1) and (2) are different and furthermore, ''exactly one'' of the two equalities holds (not both). Additional information about <math>\theta</math> is needed to determine which one holds. For example, suppose that <math>x = 0</math> and that {{em|all}} that is known about <math>\theta</math> is that <math>\,-\pi \leq \theta \leq \pi\,</math> (and nothing more is known). Then 
<math display="block">\arccos x = \arccos 0 = \frac{\pi}{2}</math> 
and moreover, in this particular case <math>k = 0</math> (for both the <math>\,+\,</math> case and the <math>\,-\,</math> case) and so consequently,
<math display="block">\theta ~=~ \pm \arccos x+2 \pi k ~=~ \pm \left(\frac{\pi}{2}\right)+2\pi (0) ~=~ \pm \frac{\pi}{2}.</math>
This means that <math>\theta</math> could be either <math>\,\pi/2\,</math> or <math>\,-\pi/2.</math> Without additional information it is not possible to determine which of these values <math>\theta</math> has. 
An example of some additional information that could determine the value of <math>\theta</math> would be knowing that the angle is above the <math>x</math>-axis (in which case <math>\theta = \pi/2</math>) or alternatively, knowing that it is below the <math>x</math>-axis (in which case <math>\theta =-\pi/2</math>).