Inverse Trigonometric Functions Lesson

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. 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, ∞). 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): $$f^{-1}(x)=\sqrt{x}$$

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. 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:

$$y=\text{sin}(x), -\frac{π}{2}≤ x ≤ \frac{π}{2}$$ If we reflect the 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)

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. $$y=\text{sin}\hspace{.1em}\frac{5π}{4}=-\frac{\sqrt{2}}{2}$$ $$y=\text{sin}\hspace{.1em}\frac{7π}{4}=-\frac{\sqrt{2}}{2}$$ 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. $$y=\text{sin}^{-1}\left(-\frac{\sqrt{2}}{2}\right)=-\frac{π}{4}$$

Cancellation Properties with Sine and Inverse Sine

$$\text{sin}(\text{sin}^{-1}(x))=x$$ $$\text{for}$$ $$-1 ≤ x ≤ 1$$ $$\text{sin}^{-1}(\text{sin}(x))=x$$ $$\text{for}$$ $$-\frac{π}{2}≤ x ≤ \frac{π}{2}$$ Example #3: Find each value. $$\text{sin}^{-1}\left(\text{sin}\left(-\frac{π}{3}\right)\right)$$ Since $-\frac{π}{3}$ is in the interval $[-\frac{π}{2}, \frac{π}{2}]$, we can use the cancellation properties of inverse sine functions. $$\text{sin}^{-1}\left(\text{sin}\left(-\frac{π}{3}\right)\right)=-\frac{π}{3}$$ Example #4: Find each value. $$\text{sin}^{-1}\left(\text{sin}\left(\frac{5π}{6}\right)\right)$$ Since $\frac{5π}{6}$ is not in the interval $[-\frac{π}{2}, \frac{π}{2}]$, we need to first evaluate the sine of $\frac{5π}{6}$. $$\text{sin}\left(\frac{5π}{6}\right)=\frac{1}{2}$$ $$\text{sin}^{-1}\left(\text{sin}\left(\frac{5π}{6}\right)\right)=\text{sin}^{-1}\left(\frac{1}{2}\right)=\frac{π}{6}$$

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. 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

$$y=\text{cos}(x), 0 ≤ x ≤ π$$ If we reflect the 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)

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 #5: 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 #6: 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. $$y=\text{cos}^{-1}\left(-\frac{\sqrt{2}}{2}\right)=\frac{3π}{4}$$

Cancellation Properties with Cosine and Inverse Cosine

$$\text{cos}(\text{cos}^{-1}(x))=x$$ $$\text{for}$$ $$-1 ≤ x ≤ 1$$ $$\text{cos}^{-1}(\text{cos}(x))=x$$ $$\text{for}$$ $$0 ≤ x ≤ π$$ Example #7: Find each value. $$\text{cos}^{-1}\left(\text{cos}\left(\frac{7π}{4}\right)\right)$$ Since $\frac{7π}{4}$ is not in the interval $[0, π]$, we need to first evaluate the cosine of $\frac{7π}{4}$. $$\text{cos}\left(\frac{7π}{4}\right)=\frac{\sqrt{2}}{2}$$ $$\text{cos}^{-1}\left(\text{cos}\left(\frac{7π}{4}\right)\right)=\text{cos}^{-1}\left(\frac{\sqrt{2}}{2}\right)=\frac{π}{4}$$

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. 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

$$y=\text{tan}(x), -\frac{π}{2}< x < \frac{π}{2}$$ If we reflect the 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)

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 #8: 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. $$y=\text{tan}^{-1}\left(\sqrt{3}\right)=\frac{π}{3}$$ Example #9: Find y in each equation.
$$y=\text{tan}^{-1}(-\sqrt{3})$$ If we think about the tangent function over our given interval of $\left(-\frac{π}{2}, \frac{π}{2}\right)$: $$y=\text{tan}\hspace{.1em}\left(-\frac{π}{3}\right)=-\sqrt{3}$$ 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. $$y=\text{tan}^{-1}\left(-\sqrt{3}\right)=-\frac{π}{3}$$

Cancellation Properties with Tangent and Inverse Tangent

$$\text{tan}(\text{tan}^{-1}(x))=x$$ $$\text{for}$$ $$x ∈ \mathbb{R}$$ $$\text{tan}^{-1}(\text{tan}(x))=x$$ $$\text{for}$$ $$-\frac{π}{2}< x < \frac{π}{2}$$ Example #10: Find each value. $$\text{tan}^{-1}\left(\text{tan}\left(\frac{4π}{3}\right)\right)$$ Since $\frac{4π}{3}$ is not in the interval $\left(-\frac{π}{2}, \frac{π}{2}\right)$, we need to first evaluate the tangent of $\frac{4π}{3}$. $$\text{tan}\left(\frac{4π}{3}\right)=\sqrt{3}$$ $$\text{tan}^{-1}\left(\text{tan}\left(\frac{4π}{3}\right)\right)=\text{tan}^{-1}\left(\sqrt{3}\right)=\frac{π}{3}$$

Inverse Cotangent, Secant, and Cosecant Functions

We can use a similar thought process to define the remaining inverse trigonometric functions. Note: These functions may be derived using an alternative approach. The graph in your textbook may not match what is given here.

Inverse Cotangent Function

$$y=\text{cot}^{-1}(x)$$ $$x=\text{cot}\hspace{.1em}y$$ $$0 < y < π$$

Inverse Secant Function

$$y=\text{sec}^{-1}(x)$$ $$x=\text{sec}\hspace{.1em}y$$ $$0 ≤ y ≤ π, y ≠ \frac{π}{2}$$

Inverse Cosecant Function

$$y=\text{csc}^{-1}(x)$$ $$x=\text{csc}\hspace{.1em}y$$ $$-\frac{π}{2}≤ y ≤ \frac{π}{2}, y ≠ 0$$ Let's look at a few examples.
Example #11: Find y in each equation.
$$y=\text{cot}^{-1}\left(-\sqrt{3}\right)$$ If we think about the cotangent function over our given interval of $\left(0, π\right)$: $$y=\text{cot}\hspace{.1em}\frac{5π}{6}=-\sqrt{3}$$ Note: This is best found from special triangles. This means the point: $\left(\frac{5π}{6}, -\sqrt{3}\right)$ is on our graph. When we work with the inverse cotangent function the points will be reversed. This tells us that: $\left(-\sqrt{3}, \frac{5π}{6}\right)$ will be on the graph of our inverse cotangent function. $$y=\text{cot}^{-1}\left(-\sqrt{3}\right)=\frac{5π}{6}$$ Example #12: Find y in each equation.
$$y=\text{sec}^{-1}\left(\sqrt{2}\right)$$ If we think about the secant function over our given interval of $\left[0, π\right], x ≠ \frac{π}{2}$: $$y=\text{sec}\hspace{.1em}\frac{π}{4}=\sqrt{2}$$ Note: This is best found from special triangles. This means the point: $\left(\frac{π}{4}, \sqrt{2}\right)$ is on our graph. When we work with the inverse secant function the points will be reversed. This tells us that: $\left(\sqrt{2}, \frac{π}{4}\right)$ will be on the graph of our inverse secant function. $$y=\text{sec}^{-1}\left(\sqrt{2}\right)=\frac{π}{4}$$ Example #13: 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 $\left[-\frac{π}{2}, \frac{π}{2}\right], x ≠ 0$: $$y=\text{csc}\hspace{.1em}\frac{5π}{3}=-\frac{2\sqrt{3}}{3}$$ 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

Composition of Inverse Trigonometric Functions

Let's look at a few problems that involve finding the composition of inverse trigonometric functions.
Example #14: 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 #15: 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. $$\text{sin}\hspace{.1em}θ=\frac{y}{r}=\frac{3}{4}$$ $$y=3, r=4$$ $$\text{tan}\hspace{.1em}θ=\frac{y}{x}$$ 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.
Example #16: Find the exact value. $$\text{sin}\left(\text{sin}^{-1}\left(\frac{1}{2}\right) + \text{tan}^{-1}(-3)\right)$$ 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: $$\frac{\sqrt{10}}{20}+ \frac{-3\sqrt{30}}{20}$$ $$\frac{\sqrt{10}- 3\sqrt{30}}{20}$$ $$\text{sin}\left(\text{sin}^{-1}\left(\frac{1}{2}\right) + \text{tan}^{-1}(-3)\right)=\frac{\sqrt{10}- 3\sqrt{30}}{20}$$

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