Hoofdstuk 11 Goniometrie *: Goniometrie in de rechthoekige driehoek *
Inverse trigonometric functions
The functions #\cos#, #\sin#, and #\tan# are not injective. But by restricting them to a suitable domain they do become so. After such a restriction, we can study the inverse function of these trigonometric functions. In this table the domains of definition and the ranges where the inverse functions exist are indicated.
| function | definition area | range | inverse function |
| #\sin# | #\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]# | #[-1,1]# | #\arcsin# |
| #\cos# | #\left[0,\pi\right]# | #[-1,1]# | #\arccos# |
| #\tan# | #\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)# | #(-\infty,\infty)\ \ # | #\arctan# |
There are more choices for the domain where we can restrict ourselves to get to an injective function. For example we can move the intervals over #2\pi#.
We avoid the notation #\sin^{-1}# for the inverse function of #\sin# on #\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]# in order to avoid confusion with the function #\dfrac{1}{\sin(x)}#.
The graphs above, of arcsine and arccosine show a strong relation. Here is the explanation:
If #-1\le x\le 1#, then #\arccos(x) = \dfrac{\pi}{2}-\arcsin(x)#.
This follows from the law of periodicity #\cos\left(\dfrac{\pi}{2}-y\right) = \sin(y)# by taking #x=\sin(y)#. We substitute #\arcsin(x)# for #y# in this law and find #\cos\left(\dfrac{\pi}{2}-\arcsin(x)\right) = x#. If we subsequently take #\arccos# to the left and the right hand side, we find #\dfrac{\pi}{2}-\arcsin(x) = \arccos(x)#, which is the equality we are proving.
Compositions of two trigonometric functions like #\cos\circ\arcsin# and #\tan\circ\arcsin# can be written without using trigonometric functions. Below are the most important examples.
Composition of a trigonometric function and an inverse trigonometric function
\[\begin{array}{rcl}\cos(\arcsin(x))&=&\sqrt{1-x^2}\\ \tan(\arcsin(x)) &=&\frac{x}{\sqrt{1-x^2}}\\ \cos(\arctan(x))&=&\frac{1}{\sqrt{x^2+1}}\\ \sin(\arctan(x))&=&\frac{x}{\sqrt{x^2+1}}\end{array}\]
If we substitute #y=\arcsin(x)# in #\sin(y)^2+\cos(y)^2=1#, and use #\arcsin(\sin(x))=x#, we find #x^2+\cos(\arcsin(x))^2=1#, from which the first equality is easily derived.
The first equality implies \[\tan(\arcsin(x)) = \frac{\sin(\arcsin(x))}{\cos(\arcsin(x))} = \frac{x}{\sqrt{1-x^2}}\tiny,\]which gives us the second equality.
For the third equality we focus on proving the third equality. For an arbitrary number#y# that is not of the form #\frac{(2k+1)\pi}{2}# for an integer #k#, \[\frac{1}{\sqrt{1+\tan(y)^2}} =\cos(y)\tiny\] applies. This can be seen by multiplying numerator and denominator of the left handside by #\cos(y)#; for the denominator then \[\begin{array}{rcl}\cos(y)\cdot\sqrt{1+\tan(y)^2} &=&\sqrt{\cos(y)^2+\cos(y)^2\cdot\frac{\sin(y)^2}{\cos(y)^2}} \\ &=& \sqrt{\cos(y)^2+\sin(y)^2}\\ &=&\sqrt{1}\\ &=&1\end{array}\] applies. Since #\cos(y)\ne0# (because of the choice of #y#), the formula #\frac{1}{\sqrt{1+\tan(y)^2}} =\cos(y)# mentioned above is proved. If we subsequently substitute #y# by #\arctan(x)#; the left hand side becomes #\frac{1}{\sqrt{1+\tan(\arctan(x))^2}}#, which is equal to #\frac{1}{\sqrt{1+x^2}}# since #\arctan# is the inverse function of #\tan#, and the rigt hand side becomes #\cos(\arctan(x))#. This proves the equality #\cos(\arctan(x))=\frac{1}{\sqrt{x^2+1}}#.
The fourth equality can be shown to hold in a similar manner.
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