Inverse Trigonometric Function Problems: All Mixed Up

This section gives mixed practice with the three most common inverse trigonometric functions:

• arcsine
• arccosine
• arctangent

All problems should be done without a calculator.

Example 1

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’:

Step 1

As shown above, 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

$\arcsin(\sin 210^\circ) = -30^\circ$

Example 2

Find:  $\,\tan(\arcsin \frac 45)$

Solution

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:

• Use the right triangle approach only: sine is opposite over hypotenuse
• Draw a right triangle with an acute angle labeled $\,\arcsin \frac 45$
• Mark lengths on the opposite side and hypotenuse, using: $\,\text{sine} = \frac 45 = \frac{\text{OPP}}{\text{HYP}}$

Step 3

Compute the remaining side length in the triangle:

• Immediately recognize the special $\,\color{red}{3}{-}4{-}5\,$ right triangle
• Or, use the Pythagorean theorem:  $\,x^2 + 4^2 = 5^2\,,$  so $\,x = \color{red}{3}$

Step 4

Tangent is opposite over adjacent:

Example 3

Rewrite as an algebraic expression in $\,t\,$:   $\,\tan(\arcsin t)$

Solution

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 $\,0 \lt t \lt 1\,,$ so you can see the similarity to the prior example. Then, we'll see that the formula obtained also works for $\,t = 0\,$ and $\,-1 \lt t \lt 0\,.$

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 $\,0 \lt t \lt 1\,.$ 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:

Mark the opposite side as $\,t\,$ and the hypotenuse as $\,1\,.$

Step 3

Compute the remaining side length in the triangle:

Since $\,x \gt 0\,$ in this initial discussion, choose the ‘$+$’ sign.

Step 4

Tangent is opposite over adjacent: 

Step 5

Note that the formula also works for $\,t = 0\,,$ as follows:

• $\tan (\arcsin 0) = \tan 0 = 0$
• For $\,t = 0\,$:  $\frac{t}{\sqrt{1-t^2}} = \frac{0}{\sqrt{1 - 0^2}} = 0$

Step 6

Note that the formula also works for $\,-1 \lt t \lt 0\,,$ as follows:

• In this case, $\,\arcsin t\,$ is in Quadrant IV.
• In Quadrant IV, the tangent is negative.
• The size of $\,\tan (\arcsin t)\,$ is the same as in Quadrant I.
• So, we want a formula that has the same size as $\,\frac{t}{\sqrt{1-t^2}}\,$ and is negative.
• Note that when $\,t\,$ is negative, so is $\,\frac{t}{\sqrt{1 - t^2}}\,$ (since it has a negative numerator and a positive denominator).

Together, $\,\tan (\arcsin t) = \frac{t}{\sqrt{1-t^2}}\,$ for $\,-1 \lt t \lt 0\,.$

Step 7

Combining all the cases:

If You Only Need a Reliable Answer ...

Zip up to WolframAlpha and type in:

arcsin ( sin(210 deg) )

tan( arcsin (4/5) )

tan( arcsin t )