Trigonometry: Trigonometric functions
Inverse trigonometric functions
We have seen that the sine, cosine and tangent are periodic functions. Therefore, if we want to solve the equation #\sin(x)=\tfrac{1}{2}#, we will find infinitely many solutions. Now we will limit the domain of the functions so that we can define an inverse function. This inverse function can help us solve equations.
We define the inverse functions of sine, cosine and tangent as follows:
\[\begin{array}{rcl} \blue x=\arcsin(\green y) &\Leftrightarrow& \green y=\sin(\blue x) \;\text{ and }\; \frac{\pi}{2} \leq \blue x \leq \frac{\pi}{2} \\ \\ \blue x=\arccos(\green y) &\Leftrightarrow& \green y=\cos(\blue x) \;\text{ and }\; 0 \leq \blue x \leq \pi \\ \\ \blue x=\arctan(\green y) &\Leftrightarrow& \green y=\tan(\blue x) \;\text{ and }\; \frac{\pi}{2} \lt \blue x \lt \frac{\pi}{2}\end{array}\]
Example
#\arcsin\left(\green{\frac{\sqrt{2}}{2}}\right) = \blue{\frac{\pi}{4}}#
because
#\sin\left(\blue{\frac{\pi}{4}}\right)=\green{\frac{\sqrt{2}}{2}}# and #\frac{\pi}{2} \leq \blue{\frac{\pi}{4}} \leq \frac{\pi}{2}#
The function #f(x)=\arcsin(x)# has
The graph is the reflection of #f(x)=\sin(x)# on the domain #\ivcc{\frac{\pi}{2}}{\frac{\pi}{2}}# across the line #y=x#. 
Plaatje

The function #f(x)=\arccos(x)# has
The graph is the reflection of #f(x)=\cos(x)# on the domain #\ivcc{0}{\pi}# across the line #y=x#. 
plaatje

The function #f(x)=\arctan(x)# has
The graph's horizontal asymptotes are #y=\frac{\pi}{2}# and #y=\frac{\pi}{2}# and the graph is the reflection of #f(x)=\tan(x)# on the domain #\ivoo{\frac{\pi}{2}}{\frac{\pi}{2}}# across the line #y=x#. 
plaatje

Because, #\sin\left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}# and #\dfrac{\pi}{2}\leq\dfrac{\pi}{4}\leq\dfrac{\pi}{2}#.
Or visit omptest.org if jou are taking an OMPT exam.