Lesson Objectives
  • Learn about inverse trigonometric functions
  • Learn how to work with the inverse sine function
  • Learn how to work with the inverse cosine function
  • Learn how to work with the inverse tangent function
  • Learn how to evaluate the composition of trigonometric functions

How to Work with Inverse Trigonometric Functions


Note: This lesson will reference lessons from our precalculus course. Specifically, we will discuss one-to-one functions, function inverses, and function inverses with a restricted domain. Lessons #38 - #42 should be covered before going further in the tutorial.

Inverse Functions

Previously, we learned how to find the inverse of a one-to-one function. Recall that a one-to-one function is a function where for each x, there is one y and for each y, there is one x. We can use the horizontal line test, to determine if a function is one-to-one.
  • In a one-to-one function, each x-value corresponds to exactly one y-value
  • In a one-to-one function, each y-value corresponds to exactly one x-value
  • If a function f is a one-to-one function, then the inverse of f is denoted as:
    • $$f^{-1}$$
  • To find f-1(x) from f(x):
    1. Replace f(x) with y
    2. Swap x and y
    3. Solve for y
    4. Replace y with f-1(x)
  • The domain of f will become the range of f-1
  • The range of f will become the domain of f-1
  • This tells us if (a, b) is on the graph of f, then (b, a) is on the graph of f-1
  • The graphs of f and f-1 are reflections across the line y = x

Finding the Domain of a Function with a Restricted Domain

If we look at the graph of the squaring function, we can see it is not a one-to-one function. Graph of the squaring function f(x)=x^2, looks like a u shape We can restrict the domain, however, to create a function that is one-to-one without changing the range. Let's suppose we choose to restrict the domain to the interval: [0, ∞). graphing f(x)=x^2 We can now find the inverse as: $$f(x)=x^2, x ≥ 0$$ Write f(x) as y: $$y=x^2, x ≥ 0$$ Swap x and y: $$x=y^2, y ≥ 0$$ Solve for y: $$y=\pm\sqrt{x}, y ≥ 0$$ Since we restricted our domain of f(x) = x2 to 0 or larger, we can throw out the negative. Recall the domain of f becomes the range of f-1. Here f-1 or y will be non-negative. $$y=\sqrt{x}$$ Replace y with f-1(x): $$f^{-1}(x)=\sqrt{x}$$ Graphing f(x)=x^2 and f^-1(x)=sqrt(x)

Inverse Sine Function

If we revisit the graph of our sine function, we can clearly see that y = sin x is not a one-to-one function. Graphing the sine function If we restrict our domain to the interval: $$\left[-\frac{π}{2}, \frac{π}{2}\right]$$ The restricted function is a one-to-one function with the same range of [-1, 1]. Since the range of sin(x) is [-1, 1], the domain of the inverse will be [-1, 1] and its range will be: $\left[-\frac{π}{2}, \frac{π}{2}\right]$

Graph of the Restricted Sine Function:

$$y=\text{sin}(x), -\frac{π}{2}≤ x ≤ \frac{π}{2}$$ graphing y=sin(x), -pi/2 <=x <=pi/2 If we reflect this graph of y = sin(x) on the restricted domain across the line y = x, we obtain the graph of our inverse sine function.

Graph of y = sin-1(x)

graph of y=sin^-1(x), also known as y=arcsin x We often see y = arcsin x instead of y = sin-1(x). We can think of y here as the number (angle) in the interval: $\left[-\frac{π}{2}, \frac{π}{2}\right]$, whose sine is x. Therefore, we often write y = sin-1(x) as sin y = x in order to evaluate it. Let's look at an example.
Example #1: Find y in each equation. $$y=\text{sin}^{-1}\left(\frac{1}{2}\right)$$ If we think about the sine function over our given interval of $\left[-\frac{π}{2}, \frac{π}{2}\right]$: $$y=\text{sin}\hspace{.1em}\frac{π}{6}=\frac{1}{2}$$ $$\frac{1}{2}=\text{sin}\hspace{.1em}\frac{π}{6}$$ This means the point: $\left(\frac{π}{6}, \frac{1}{2}\right)$ is on our graph. When we work with the inverse sine function the points will be reversed. This tells us that: $\left(\frac{1}{2}, \frac{π}{6}\right)$ will be on the graph of our inverse sine function. $$y=\text{sin}^{-1}\left(\frac{1}{2}\right)=\frac{π}{6}$$ Example #2: Find y in each equation. $$y=\text{sin}^{-1}\left(-\frac{\sqrt{2}}{2}\right)$$ With this example, we must be careful of how we give our answer. We know that sine is negative in quadrants III and IV. $$y=\text{sin}\hspace{.1em}\frac{7π}{4}=-\frac{\sqrt{2}}{2}$$ $$y=\text{sin}\hspace{.1em}\frac{5π}{4}=-\frac{\sqrt{2}}{2}$$ Notice that neither value is in the domain of the restricted sine function. Recall that if we rotate clockwise, we obtain a negative angle measure. If we simply add $-2π$ to our angle measure in quadrant IV, we will be within our domain and still have the same sine value. $$\frac{7π}{4}- \frac{8π}{4}=-\frac{π}{4}$$ We can write this as: $$y=\text{sin}\hspace{.1em}\left(-\frac{π}{4}\right)=-\frac{\sqrt{2}}{2}$$ Let's use this to answer our problem. $$y=\text{sin}^{-1}\left(-\frac{\sqrt{2}}{2}\right)=-\frac{π}{4}$$

Inverse Cosine Function

We have previously worked with the graph of the cosine function. Just like the sine function, we can clearly see that y = cos x is not a one-to-one function. Graphing the cosine function If we restrict our domain to the interval: $$\left[0, π\right]$$ The restricted function is a one-to-one function with the same range of [-1, 1]. Since the range of cos(x) is [-1, 1], the domain of the inverse will be [-1, 1] and its range will be [0, $π$].

Graph of the Restricted Cosine Function

$$y=\text{cos}(x), 0 ≤ x ≤ π$$ Graphing the cosine function with a restricted domain If we reflect this graph of y = cos(x) on the restricted domain across the line y = x, we obtain the graph of our inverse cosine function.

Graph of y = cos-1(x)

Graphing the inverse cosine function We often see y = arccos x instead of y = cos-1(x). We can think of y here as the number (angle) in the interval: $\left[0, π\right]$, whose cosine is x. Therefore, we often write y = cos-1(x) as cos y = x in order to evaluate it. Let's look at an example.
Example #3: Find y in each equation.
$$y=\text{cos}^{-1}\left(-\frac{\sqrt{3}}{2}\right)$$ If we think about the cosine function over our given interval of $\left[0, π\right]$: $$y=\text{cos}\left(\frac{5π}{6}\right)=-\frac{\sqrt{3}}{2}$$ $$-\frac{\sqrt{3}}{2}=\text{cos}\left(\frac{5π}{6}\right)$$ This means the point: $\left(\frac{5π}{6}, -\frac{\sqrt{3}}{2}\right)$ is on our graph. When we work with the inverse cosine function the points will be reversed. This tells us that: $\left(-\frac{\sqrt{3}}{2}, \frac{5π}{6}\right)$ will be on the graph of our inverse cosine function. $$y=\text{cos}^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\frac{5π}{6}$$ Example #4: Find y in each equation.
$$y=\text{cos}^{-1}\left(-\frac{\sqrt{2}}{2}\right)$$ If we think about the cosine function over our given interval of $\left[0, π\right]$: $$y=\text{cos}\left(\frac{3π}{4}\right)=-\frac{\sqrt{2}}{2}$$ $$-\frac{\sqrt{2}}{2}=\text{cos}\left(\frac{3π}{4}\right)$$ This means the point: $\left(\frac{3π}{4}, -\frac{\sqrt{2}}{2}\right)$ is on our graph. When we work with the inverse cosine function the points will be reversed. This tells us that: $\left(-\frac{\sqrt{2}}{2}, \frac{3π}{4}\right)$ will be on the graph of our inverse cosine function. $$y=\text{cos}^{-1}\left(-\frac{\sqrt{2}}{2}\right)=\frac{3π}{4}$$

Inverse Tangent Function

Lastly, if we revisit the graph of our tangent function, we can clearly see that y = tan x is not a one-to-one function. Graphing the tangent function If we restrict our domain to the interval: $$\left(-\frac{π}{2}, \frac{π}{2}\right)$$ The restricted function is a one-to-one function with the same range of (-∞, ∞). Since the range of tan(x) is (-∞, ∞), the domain of the inverse will be (-∞, ∞) and its range will be $\left(-\frac{π}{2}, \frac{π}{2}\right)$.

Graph of the Restricted Tangent Function

$$y=\text{tan}(x), -\frac{π}{2}≤ x ≤ \frac{π}{2}$$ Graphing the tangent function If we reflect this graph of y = tan(x) on the restricted domain across the line y = x, we obtain the graph of our inverse tangent function.

Graph of y = tan-1(x)

Sketching the graph of the inverse tangent function We often see y = arctan x instead of y = tan-1(x). We can think of y here as the number (angle) in the interval: $\left(-\frac{π}{2}, \frac{π}{2}\right)$, whose tangent is x. Therefore, we often write y = tan-1(x) as tan y = x in order to evaluate it. Let's look at an example.
Example #5: Find y in each equation.
$$y=\text{tan}^{-1}(\sqrt{3})$$ If we think about the tangent function over our given interval of $\left(-\frac{π}{2}, \frac{π}{2}\right)$: $$y=\text{tan}\hspace{.1em}\frac{π}{3}=\sqrt{3}$$ Note: This is best found from special triangles. $$\sqrt{3}=\text{tan}\left(\frac{π}{3}\right)$$ This means the point: $\left(\frac{π}{3}, \sqrt{3}\right)$ is on our graph. When we work with the inverse tangent function the points will be reversed. This tells us that: $\left(\sqrt{3}, \frac{π}{3}\right)$ will be on the graph of our inverse tangent function. $$y=\text{tan}^{-1}\left(\sqrt{3}\right)=\frac{π}{3}$$

Inverse Cotangent, Secant, and Cosecant Functions

Inverse Cotangent Function: $$0 < y < π$$ $$y=\text{cot}^{-1}(x)$$ $$x=\text{cot}\hspace{.1em}y$$ Inverse Secant Function: $$0 ≤ y ≤ π, y ≠ \frac{π}{2}$$ $$y=\text{sec}^{-1}(x)$$ $$x=\text{sec}\hspace{.1em}y$$ Inverse Cosecant Function: $$-\frac{π}{2}≤ y ≤ \frac{π}{2}, y ≠ 0$$ $$y=\text{csc}^{-1}(x)$$ $$x=\text{csc}\hspace{.1em}y$$ Let's look at an example.
Example #6: Find y in each equation.
$$y=\text{csc}^{-1}\left(-\frac{2\sqrt{3}}{3}\right)$$ If we think about the cosecant function over our given interval of $\left[-\frac{π}{2}, \frac{π}{2}\right], y ≠ 0$: $$y=\text{csc}\hspace{.1em}\frac{5π}{3}=-\frac{2\sqrt{3}}{3}$$ Note: This is best found from special triangles. Since this is not in our given interval, we need to add -2$π$ $$\frac{5π}{3}- \frac{6π}{3}=-\frac{π}{3}$$ $$y=\text{csc}\hspace{.1em}\left(-\frac{π}{3}\right)=-\frac{2\sqrt{3}}{3}$$ This means the point: $\left(-\frac{π}{3}, -\frac{2\sqrt{3}}{3}\right)$ is on our graph. When we work with the inverse cosecant function the points will be reversed. This tells us that: $\left(-\frac{2\sqrt{3}}{3}, -\frac{π}{3}\right)$ will be on the graph of our inverse cosecant function. $$y=\text{csc}^{-1}\left(-\frac{2\sqrt{3}}{3}\right)=-\frac{π}{3}$$

Summary Table of Inverse Trigonometric Functions

Inverse Function Domain Range Unit Circle
$y=\text{sin}^{-1}(x)$$[-1, 1]$$\left[-\frac{π}{2}, \frac{π}{2}\right]$I and IV
$y=\text{cos}^{-1}(x)$$[-1, 1]$$[0, π]$I and II
$y=\text{tan}^{-1}(x)$$(-\infty, \infty)$$\left(-\frac{π}{2}, \frac{π}{2}\right)$I and IV
$y=\text{cot}^{-1}(x)$$(-\infty, \infty)$$\left(0, π\right)$I and II
$y=\text{sec}^{-1}(x)$$(-\infty, -1] ∪ [1, \infty)$$\left[0, \frac{π}{2}\right) ∪ \left(\frac{π}{2}, π\right]$I and II
$y=\text{csc}^{-1}(x)$$(-\infty, -1] ∪ [1, \infty)$$\left[-\frac{π}{2}, 0\right) ∪ (0, \frac{π}{2}]$I and IV

Composition of Inverse Trigonometric Functions

Let's look at a few problems that involve finding the composition of inverse trigonometric functions.
Example #7: Find the exact value. $$\text{sin}^{-1}\left(\text{tan}\left(-\frac{π}{4}\right)\right)$$ For these types of problems, we start with the inside function. $$\text{tan}\left(-\frac{π}{4}\right)=-1$$ Let's replace this in our original problem: $$\text{sin}^{-1}(-1)=-\frac{π}{2}$$ Example #8: Find the exact value. $$\text{tan}\left(\text{sin}^{-1}\left(\frac{3}{4}\right)\right)$$ These types of problems can be a bit tough to understand at first. Let's first consider the following: $$\text{sin}\hspace{.1em}θ=\frac{3}{4}$$ $$\text{sin}^{-1}\hspace{.1em}\left(\frac{3}{4}\right)=θ$$ Let's now make a little substitution: $$\text{tan}\left(\text{sin}^{-1}\left(\frac{3}{4}\right)\right)=\text{tan}\hspace{.1em}θ$$ Since the sine of our unknown angle θ is positive, θ can only be in quadrants I or II.
The result of our inverse sine function will give us an angle measure that is in quadrant I or quadrant IV. Therefore, we know that θ lies in quadrant I.
Earlier in the course, we learned how to find trigonometric function values given one ratio and the quadrant. We know that: $$\text{sin}\hspace{.1em}θ=\frac{y}{r}=\frac{3}{4}$$ $$y=3, r=4$$ We know that: $$\text{tan}\hspace{.1em}θ=\frac{y}{x}$$ We can find x using the Pythagorean Formula: $$x^2 + y^2=r^2$$ $$x^2 + 3^2=4^2$$ $$x^2 + 9=16$$ $$x^2=7$$ Take the principal square root since we are in quadrant I. $$x=\sqrt{7}$$ $$\text{tan}\hspace{.1em}θ=\frac{y}{x}=\frac{3}{\sqrt{7}}$$ Rationalize the Denominator: $$\text{tan}\hspace{.1em}θ=\frac{3}{\sqrt{7}}\cdot \frac{\sqrt{7}}{\sqrt{7}}=\frac{3\sqrt{7}}{7}$$ Now, we can put this answer in terms of our original problem: $$\text{tan}\left(\text{sin}^{-1}\left(\frac{3}{4}\right)\right)=\frac{3\sqrt{7}}{7}$$

Finding the Exact Value Using Identities

Lastly, let's look at a problem that involves using an identity. $$\text{sin}\left(\text{sin}^{-1}\left(\frac{1}{2}\right) + \text{tan}^{-1}(-3)\right)$$ Let's replace sin-1(1/2) with $π$/6: $$\text{sin}\left(\frac{π}{6}+ \text{tan}^{-1}(-3)\right)$$ Since we don't know the value of tan-1(-3), let's just set this equal to β for now: $$\text{tan}^{-1}(-3)=β$$ $$\text{tan}\hspace{.1em}β=-3$$ Let's replace tan-1(-3) with $β$: $$\text{sin}\left(\frac{π}{6}+ β\right)$$ Recall the sum identity for sine: $$\text{sin}(A + B)=\text{sin}\hspace{.1em}A \hspace{.1em}\text{cos}B + \text{cos}\hspace{.1em}A \hspace{.1em}\text{sin}\hspace{.1em}B$$ Let's use this identity to rewrite our problem: $$\text{sin}\left(\frac{π}{6}+ β\right)$$ $$\text{sin}\hspace{.1em}\frac{π}{6}\hspace{.1em}\text{cos}β + \text{cos}\hspace{.1em}\frac{π}{6}\hspace{.1em}\text{sin}\hspace{.1em}β$$ $$\text{sin}\hspace{.1em}\frac{π}{6}=\frac{1}{2}$$ $$\text{cos}\hspace{.1em}\frac{π}{6}=\frac{\sqrt{3}}{2}$$ Let's replace these in our problem: $$\frac{1}{2}\text{cos}β + \frac{\sqrt{3}}{2}\hspace{.1em}\text{sin}\hspace{.1em}β$$ Now, let's think about how we can find the cos β and the sin β $$\text{tan}^{-1}(-3)=β$$ $$\text{tan}\hspace{.1em}β=-3$$ Tangent is negative in quadrants II and IV. Since the result of our inverse tangent function gives us an angle in quadrant I or IV, we know that our angle β is in quadrant IV.
Here, x-values will be positive and y-values will be negative. $$\text{tan}\hspace{.1em}β=\frac{y}{x}=\frac{-3}{1}$$ $$x=1, y=-3$$ Find r: $$x^2 + y^2=r^2$$ $$1 + 9=r^2$$ $$r^2=10$$ $$r=\sqrt{10}$$ Remember r is always positive.
Let's find cos β: $$\text{cos}\hspace{.1em}β=\frac{x}{r}=\frac{1}{\sqrt{10}}$$ Rationalize the Denominator: $$\text{cos}\hspace{.1em}β=\frac{1}{\sqrt{10}}\cdot \frac{\sqrt{10}}{\sqrt{10}}=\frac{\sqrt{10}}{10}$$ Let's find sin β: $$\text{sin}\hspace{.1em}β=\frac{y}{r}=\frac{-3}{\sqrt{10}}$$ Rationalize the Denominator: $$\text{sin}\hspace{.1em}β=\frac{-3}{\sqrt{10}}\cdot \frac{\sqrt{10}}{\sqrt{10}}=\frac{-3\sqrt{10}}{10}$$ Let's replace cos β and sin β: $$\frac{1}{2}\text{cos}β + \frac{\sqrt{3}}{2}\hspace{.1em}\text{sin}\hspace{.1em}β$$ $$\frac{1}{2}\cdot \frac{\sqrt{10}}{10}+ \frac{\sqrt{3}}{2}\cdot \frac{-3\sqrt{10}}{10}$$ Simplify: $$\frac{\sqrt{10}}{20}+ \frac{-3\sqrt{30}}{20}$$ $$\frac{\sqrt{10}- 3\sqrt{30}}{20}$$ $$\text{sin}\left(\text{sin}^{-1}\left(\frac{1}{2}\right) + \text{tan}^{-1}(-3)\right)=\frac{\sqrt{10}- 3\sqrt{30}}{20}$$

Skills Check:

Example #1

Find y in each equation. $$y=\text{sin}^{-1}(-3)$$

Please choose the best answer.

A
$$\frac{π}{6}$$
B
$$\frac{π}{4}$$
C
$$\text{Does Not Exist}$$
D
$$-\frac{π}{6}$$
E
$$-\frac{π}{3}$$

Example #2

Find y in each equation. $$y=\text{tan}^{-1}\left(\frac{\sqrt{3}}{3}\right)$$

Please choose the best answer.

A
$$\frac{π}{6}$$
B
$$\text{Does Not Exist}$$
C
$$\frac{π}{3}$$
D
$$-\frac{π}{4}$$
E
$$-\frac{π}{6}$$

Example #3

Find the exact value. $$\text{cos}\left(\text{tan}^{-1}\left(\frac{3}{4}\right)\right)$$

Please choose the best answer.

A
$$\frac{3}{5}$$
B
$$\frac{5π}{6}$$
C
$$\frac{4}{5}$$
D
$$-\frac{π}{6}$$
E
$$\frac{\sqrt{2}}{2}$$
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