Editing Inverse trigonometric functions (section)

===Transforming equations===

The equations above can be transformed by using the reflection and shift identities:<ref>{{harvnb|Abramowitz|Stegun|1972|loc=p. 73, 4.3.44}}</ref>

{| class="wikitable" style="text-align: center;"
|+ Transforming equations by shifts and reflections
|-
! scope="col" | Argument: <math>\underline{\;~~~~~~\;}=</math>
! scope="col" |<math>-\theta</math>
! scope="col" |<math>\frac{\pi}{2} \pm \theta</math>
! scope="col" |<math>\pi \pm \theta</math>
! scope="col" |<math>\frac{3\pi}{2} \pm \theta</math>
! scope="col" |<math>2 k \pi \pm \theta,</math><br/> <math>(k \in \Z)</math>
|- <!-- sin -->
! scope="row" |<math>\sin \underline{\;~~~~~~~~~~~~~~\;}=</math>
| <math>-\sin \theta</math>
| <math>\phantom{-}\cos \theta</math>
| <math>\mp\sin \theta</math>
| <math>-\cos \theta</math>
| <math>\pm\sin \theta</math>
|- <!-- csc -->
! scope="row" |<math>\csc \underline{\;~~~~~~~~~~~~~~\;}=</math>
| <math>-\csc \theta</math>
| <math>\phantom{-}\sec \theta</math>
| <math>\mp\csc \theta</math>
| <math>-\sec \theta</math>
| <math>\pm\csc \theta</math>
|- <!-- cos -->
! scope="row" |<math>\cos \underline{\;~~~~~~~~~~~~~~\;}=</math>
| <math>\phantom{-}\cos \theta</math>
| <math>\mp\sin \theta</math>
| <math>-\cos \theta</math>
| <math>\pm\sin \theta</math>
| <math>\phantom{-}\cos \theta</math>
|- <!-- sec -->
! scope="row" |<math>\sec \underline{\;~~~~~~~~~~~~~~\;}=</math>
| <math>\phantom{-}\sec \theta</math>
| <math>\mp\csc \theta</math>
| <math>-\sec \theta</math>
| <math>\pm\csc \theta</math>
| <math>\phantom{-}\sec \theta</math>
|- <!-- tan -->
! scope="row" |<math>\tan \underline{\;~~~~~~~~~~~~~~\;}=</math>
| <math>-\tan \theta</math>
| <math>\mp\cot \theta</math>
| <math>\pm\tan \theta</math>
| <math>\mp\cot \theta</math>
| <math>\pm\tan \theta</math>
|- <!-- cot -->
! scope="row" |<math>\cot \underline{\;~~~~~~~~~~~~~~\;}=</math>
| <math>-\cot \theta</math>
| <math>\mp\tan \theta</math>
| <math>\pm\cot \theta</math>
| <math>\mp\tan \theta</math>
| <math>\pm\cot \theta</math>
|}

These formulas imply, in particular, that the following hold:

<math display=block>
\begin{align}
\sin \theta
&= -\sin(-\theta)
&&= -\sin(\pi+\theta)
&&= \phantom{-}\sin(\pi-\theta) \\
&= -\cos\left(\frac{\pi}{2}+\theta\right)
&&= \phantom{-}\cos\left(\frac{\pi}{2}-\theta\right)
&&= -\cos\left(-\frac{\pi}{2}-\theta\right) \\
&= \phantom{-}\cos\left(-\frac{\pi}{2}+\theta\right)
&&= -\cos\left(\frac{3\pi}{2}-\theta\right)
&&= -\cos\left(-\frac{3\pi}{2}+\theta\right)
\\[0.3ex]
\cos \theta
&= \phantom{-}\cos(-\theta)
&&= -\cos(\pi+\theta)
&&= -\cos(\pi-\theta) \\
&= \phantom{-}\sin\left(\frac{\pi}{2}+\theta\right)
&&= \phantom{-}\sin\left(\frac{\pi}{2}-\theta\right) 
&&= -\sin\left(-\frac{\pi}{2}-\theta\right) \\
&= -\sin\left(-\frac{\pi}{2}+\theta\right)
&&= -\sin\left(\frac{3\pi}{2}-\theta\right)
&&= \phantom{-}\sin\left(-\frac{3\pi}{2}+\theta\right)
\\[0.3ex]
\tan \theta
&= -\tan(-\theta)
&&= \phantom{-}\tan(\pi+\theta)
&&= -\tan(\pi-\theta) \\
&= -\cot\left(\frac{\pi}{2}+\theta\right)
&&= \phantom{-}\cot\left(\frac{\pi}{2}-\theta\right)
&&= \phantom{-}\cot\left(-\frac{\pi}{2}-\theta\right) \\
&= -\cot\left(-\frac{\pi}{2}+\theta\right)
&&= \phantom{-}\cot\left(\frac{3\pi}{2}-\theta\right)
&&= -\cot\left(-\frac{3\pi}{2}+\theta\right)
\\[0.3ex]
\end{align}
</math>

where swapping <math>\sin \leftrightarrow \csc,</math> swapping <math>\cos \leftrightarrow \sec,</math> and swapping <math>\tan \leftrightarrow \cot</math> gives the analogous equations for <math>\csc, \sec, \text{ and } \cot,</math> respectively.

So for example, by using the equality <math display="inline">\sin \left(\frac{\pi}{2}-\theta\right) = \cos \theta,</math> the equation <math>\cos \theta = x</math> can be transformed into <math display="inline">\sin \left(\frac{\pi}{2}-\theta\right) = x,</math> which allows for the solution to the equation <math>\;\sin \varphi = x\;</math> (where <math display="inline">\varphi := \frac{\pi}{2}-\theta</math>) to be used; that solution being: 
<math>\varphi  = (-1)^k \arcsin (x)+\pi k \; \text{ for some } k \in \Z,</math>
which becomes:
<math display="block">\frac{\pi}{2}-\theta ~=~ (-1)^k \arcsin (x)+\pi k \quad \text{ for some } k \in \Z</math>
where using the fact that <math>(-1)^{k} = (-1)^{-k}</math> and substituting <math>h :=-k</math> proves that another solution to <math>\;\cos \theta = x\;</math> is:
<math display="block">\theta ~=~ (-1)^{h+1} \arcsin (x)+\pi h+\frac{\pi}{2} \quad \text{ for some } h \in \Z.</math>
The substitution <math>\;\arcsin x = \frac{\pi}{2}-\arccos x\;</math> may be used express the right hand side of the above formula in terms of <math>\;\arccos x\;</math> instead of <math>\;\arcsin x.\;</math>