This section gives mixed practice with the three most common inverse trigonometric functions:
All problems should be done without a calculator.Compute: $\,\arcsin ( \sin 210^\circ)$
Solution:
Caution!
Since the sine and arcsine functions are not true inverses, they do not necessarily ‘undo’ each other.
Indeed, $\,\arcsin( \sin 210^\circ)\,$ cannot equal $\,210^\circ\,$,
since the arcsine always returns an angle between $\,90^\circ\,$ and $\,90^\circ\,$.
Work from the ‘inside’ of the expression to the ‘outside’:
$$\cssId{s14}{
\overset{\text{step 3}}
{\overbrace{\arcsin (\ \underset{\text{step 2}}
{ \underbrace{\sin \overset{\text{step 1}}
{\overbrace{210^\circ}}
}}\ )}}}
$$
Step 1: 
As shown at right, lay off a $\,210^\circ\,$ angle ($\,210^\circ = 180^\circ + 30^\circ\,$) in the unit circle. Recall that positive angles are laid off in a counterclockwise direction. Observe that the reference angle for $\,210^\circ\,$ is $\,\color{red}{30^\circ}\,$. 

Step 2:  By definition of the sine function, $\,\sin 210^\circ\,$ is the $y$value of the terminal point.  
Step 3: 
Apply the arcsine function: What angle between $\,90^\circ\,$ and $\,90^\circ\,$ has this $y$value? Answer: $\,30^\circ$ 

Report the answer:  $\arcsin(\sin 210^\circ) = 30^\circ$ 
Find: $\,\tan(\arcsin \frac 45)\,$
Step 1: 
Show the angle $\,\arcsin \frac{4}{5}\,$ on a unit circle.
It is the unique angle between $\,90^\circ\,$ and $\,90^\circ\,$ whose sine value is $\,\frac 45\,$; i.e., the unique angle (in the first quadrant) whose terminal point has $y$value equal to $\,\frac 45\,$. 

Step 2: 
Take the green triangle (which has hypotenuse of length $\,1\,$ and vertical side of length $\,\frac 45\,$)
and multiply all the side lengths by $\,5\,$. This gives a similar triangle with side lengths that are much easier to work with! Alternative approach for steps 1 and 2:


Step 3: 
Compute the remaining side length in the triangle:


Step 4:  Tangent is opposite over adjacent: $ \displaystyle\tan(\arcsin \frac 45) = \frac{\text{OPP}}{\text{ADJ}} = \frac 43 $ 
Rewrite as an algebraic expression in $\,t\,$: $\,\tan(\arcsin t)\,$
This is similar to the prior example, except for the variable $\,t\,$ instead of the constant $\,\frac 45\,$.
For what values of $\,t\,$ is $\,\arcsin t\,$ defined?
In other words, what is the domain of the arcsine function?
Answer: $\,t\in[1,1]\,$
Initially, we'll suppose that $\,0 < t < 1\,$, so you can see the similarity to the prior example.
Then, we'll see that the formula obtained also works for $\,t = 0\,$ and $\,1 < t < 0\,$.
Since both $\,\tan(\arcsin 1) = \tan(90^\circ)\,$ and $\,\tan(\arcsin(1)) = \tan(90^\circ)\,$ do not exist,
the expression $\,\tan(\arcsin t)\,$ is not defined at $\,t = \pm 1\,$.
Step 1: 
Let $\,0 < t < 1\,$.
Show the angle $\,\arcsin t\,$ on a unit circle. It is the unique angle between $\,90^\circ\,$ and $\,90^\circ\,$ whose sine value is $\,t\,$; i.e., the unique angle (in the first quadrant) whose terminal point has $y$value equal to $\,t\,$. 

Step 2: 
Or, use the right triangle approach:
sine is opposite over hypotenuse, so
$$
\cssId{s66}{\text{sine} = t = \frac {t}{1} = \frac{\text{OPP}}{\text{HYP}}}
$$
Mark the opposite side as $\,t\,$ and the hypotenuse as $\,1\,$.


Step 3:  Compute the remaining side length in the triangle: $$ \begin{gather} \cssId{s70}{x^2 + t^2 = 1^2}\cr \cssId{s71}{x^2 = 1  t^2}\cr \cssId{s72}{x = \pm\sqrt{1  t^2}} \end{gather} $$ Since $\,x > 0\,$ in this initial discussion, choose the ‘$+$’ sign.  
Step 4:  Tangent is opposite over adjacent: $ \displaystyle\tan(\arcsin t) = \frac{\text{OPP}}{\text{ADJ}} = \frac{t}{\sqrt{1t^2}} $  
Step 5: 
Note that the formula also works for $\,t = 0\,$, as follows:


Step 6: 
Note that the formula also works for $\,1 < t < 0\,$, as follows:


Step 7:  Combining all the cases: $$ \cssId{s92}{\tan (\arcsin t) = \frac{t}{\sqrt{1  t^2}}\ \ \ \text{for}\ \ \ 1 \lt t\lt 1} $$ 
Zip up to WolframAlpha and type in:
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
