We know that \(\sin (30^o)=\dfrac{1}{2}\). Now, suppose we are asked to find an angle,\(\theta\), whose sine is \(\dfrac{1}{2}\). We can't solve this problem with the normal trigonometric functions, we need inverse trigonometric functions! What are those?
In this article, we go over what inverse trigonometric functions are and discuss their formulas, graphs, and examples in detail. But before moving on, if you need to review inverse functions, please refer to our Inverse Functions article.
- What is an inverse trigonometric function?
- Inverse trigonometric functions: formulas
- Inverse trigonometric function graphs
- Inverse trigonometric functions: unit circle
- The calculus of inverse trigonometric functions
- Solving inverse trigonometric functions: examples
What is an Inverse Trigonometric Function?
From our Inverse Functions article, we remember that the inverse of a function can be found algebraically by switching the x- and y-values and then solving for y. We also remember that we can find the graph of the inverse of a function by reflecting the graph of the original function over the line \(y=x\).
We already know about inverse operations. For instance, addition and subtraction are inverses, and multiplication and division are inverses.
The key here is: an operation (like addition) does the opposite of its inverse (like subtraction).
In trigonometry, this idea is the same. Inverse trigonometric functions do the opposite of the normal trigonometric functions. More specifically,
Inverse sine, \(sin^{-1}\) or \(arcsin\), does the opposite of the sine function.
Inverse cosine, \(cos^{-1}\) or \(arccos\) , does the opposite of the cosine function.
Inverse tangent, \(tan^{-1}\) or \(arctan\), does the opposite of the tangent function.
Inverse cotangent, \(cot^{-1}\) or \(arccot\), does the opposite of the cotangent function.
Inverse secant, \(sec^{-1}\) or \(arcsec\), does the opposite of the secant function.
Inverse cosecant, \(csc^{-1}\) or \(arccsc\), does the opposite of the cosecant function.
The inverse trigonometric functions are also called arc functions because, when given a value, they return the length of the arc needed to obtain that value. This is why we sometimes see inverse trig functions written as \(arcsin, arccos, arctan\), etc.
Using the right triangle below, let's define the inverse trig functions!
Fig. 1. A right triangle with the sides labeled.
The inverse trigonometric functions are inverse operations to the trigonometric functions. In other words, they do the opposite of what the trig functions do. In general, if we know a trig ratio but not the angle, we can use an inverse trig function to find the angle. This leads us to define them in the following way:
| Trig functions – given an angle, return a ratio | Inverse trig functions – given a ratio, return an angle |
| \[\sin(\theta)=\dfrac{opposite}{hypotenuse}\] | \[(\theta)=sin^{-1} \dfrac{opposite}{hypotenuse}\] |
| \[\cos(\theta)=\dfrac{adjacent}{hypotenuse}\] | \[(\theta)=cos^{-1}\dfrac{adjacent}{hypotenuse}\] |
| \[\tan(\theta)=\dfrac{opposite}{adjacent}\] | \[(\theta)=\tan^{-1}\dfrac{opposite}{adjacent}\] |
| \[\cot(\theta)=\dfrac{adjacent}{opposite}\] | \[(\theta)=\cot^{-1}\dfrac{adjacent}{opposite}\] |
| \[\sec(\theta)=\dfrac{hypotenuse}{adjacent}\] | \[(\theta)=\sec^{-1}\dfrac{hypotenuse}{adjacent}\] |
| \[\csc(\theta)=\dfrac{hypotenuse}{opposite}\] | \[(\theta)=csc^{-1}\dfrac{hypotenuse}{opposite}\] |
A Note on Notation
As you might have noticed, the notation used to define the inverse trig functions makes it look like they have exponents. While it may seem like it, the \(-1\) superscript is NOT an exponent! In other words, \(\sin^{-1}(x)\) is not the same as \(\dfrac{1}{\sin(x)}\)! The \(-1\) superscript simply means “inverse.”
For perspective, if we were to raise a number or variable to the \(-1\) power, this means we are asking for its multiplicative inverse, or its reciprocal.
- For instance, \(5^{-1}=\dfrac{1}{5}\).
- And in general, if the variable is a nonzero real number, then \(c^{-1}=\dfrac{1}{c}\).
So, why are the inverse trig functions any different?
- Because inverse trig functions are functions, not quantities!
- In general, when we see a \(-1\) superscript after a function name, that means it is an inverse function, not a reciprocal!
Therefore:
- If we have a function called \(f\), then its inverse would be called \(f^{-1}\) .
- If we have a function called \(f(x)\), then its inverse would be called \(f^{-1}(x)\).
This pattern continues for any function!
Inverse Trigonometric Functions: Formulas
The main inverse trigonometric formulas are listed in the table below.
| The 6 main inverse trigonometric formulas |
| Inverse sine, or, arc sine: \(y=sin^{-1}(x)=arcsin(x)\) | Inverse cosecant, or, arc cosecant: \(y=csc^{-1}(x)=arccsc(x)\) |
| Inverse cosine, or, arc cosine: \(y=cos^{-1}(x)=arccos(x)\) | Inverse secant, or, arc secant: \(y=sec^{-1}(x)=arcsec(x)\) |
| Inverse tangent, or, arc tangent: \(y=tan^{-1}(x)=arctan(x)\) | Inverse cotangent, or, arc cotangent: \(y=cot^{-1}(x)=arcot(x)\) |
Let's explore these with an example!
Consider the inverse trigonometric function: \(y=sin^{-1}(x)\)
Based on the definition of inverse trigonometric functions, this implies that: \(sin(y)=x\).
Keeping this in mind, say we want to find the angle θ in the right triangle below. How can we go about doing so?
Fig. 2.A right triangle with its sides labeled with numbers.
Solution:
- Try using trig functions:
- We know that: \(\sin(\theta)=\dfrac{opposite}{hypotenuse}=\dfrac{1}{2}\), but this doesn't help us find the angle.
- So, what can we try next?
- Use inverse trig functions:
- Remembering the definition of inverse trig functions, if \(\sin(\theta)=\dfrac{1}{2}\), then \(\theta=\sin^{-1}\left(\dfrac{1}{2}\right)\).
- Based on our previous knowledge of trig functions, we know that \(\sin(30^o)=\dfrac{1}{2}\).
- Therefore:
- \(\theta=\sin^{-1}\left(\dfrac{1}{2}\right)\)
- \(\theta=30^o\)
Inverse Trigonometric Function Graphs
What do the inverse trigonometric functions look like? Let's check out their graphs.
Domain and Range of Inverse Trigonometric Functions
But, before we can graph the inverse trigonometric functions, we need to talk about their domains. Because the trigonometric functions are periodic, and therefore not one-to-one, they don't have inverse functions. So then, how can we have inverse trigonometric functions?
To find inverses of the trigonometric functions, we must either restrict or specify their domains so that they are one-to-one! Doing so allows us to define a unique inverse of either sine, cosine, tangent, cosecant, secant, or cotangent.
In general, we use the following convention when evaluating inverse trigonometric functions:
| Inverse trig function | Formula | Domain |
| Inverse sine / arc sine | \(y=sin^{-1}(x)=arcsin(x)\) | \([-1,1]\) |
| Inverse cosine / arc cosine | \(y=cos^{-1}(x)=arccos(x)\) | \([-1,1]\) |
| Inverse tangent / arc tangent | \(y=tan^{-1}(x)=arctan(x)\) | \(-\infty, \infty\) |
| Inverse cotangent / arc cotangent | \(y=cot^{-1}(x)=arccot(x)\) | \(-\infty, infty\) |
| Inverse secant / arc secant | \(y=sec^{-1}(x)=arcsec(x)\) | \((-\infty, -1] \cup [1, \infty)\) |
| Inverse cosecant / arc cosecant | \(y=csc^{-1}(x)=arccsc(x)\) | \((-\infty, -1] \cup [1, \infty)\) |
These are just the conventional, or standard, domain we choose when restricting the domains. Remember, since trig functions are periodic, there are an infinite number of intervals on which they are one-to-one!
To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains specified in the table above and reflect those graphs about the line \(y=x\), just as we did for finding Inverse Functions.
Below are the 6 main inverse trigonometric functions and their graphs, domain, range (also known as the principal interval), and any asymptotes.
| The graph of \(y=sin^{-1}(x)=arcsin(x)\) | The graph of \(y=cos^{-1}(x)=arccos(x)\) |
| 
|
| Domain: \([-1,1]\) | Range: \([-\dfrac{\pi}{2},\dfrac{\pi}{2}]\) | Domain: \([-1,1]\) | Range: \([0,\pi]\) |
| The graph of \(y=sec^{-1}(x)=arcsec(x)\) | The graph of \(y=csc^{-1}(x)=arccsc(x)\) |
| 
|
| Domain: \((-\infty, -1] \cup [1, \infty)\) | Range: \((0, \dfrac{\pi}{2}] \cup [\dfrac{\pi}{2}, \pi)\) | Domain: \((-\infty, -1] \cup [1, \infty)\) | Range: \((- \dfrac{\pi}{2},0] \cup [0,\dfrac{\pi}{2})\) |
| Asymptote: \(y=\dfrac{\pi}{2}\) | Asymptote: \(y=0\) |
| The graph of \(y=tan^{-1}(x)=arctan(x)\) | The graph of \(y=cot^{-1}(x)=arccot(x)\) |
| 
|
| Domain: \(-\infty, \infty\) | Range: \([-\dfrac{\pi}{2},\dfrac{\pi}{2}]\) | Domain: \(-\infty, \infty\) | Range: \(0, \pi\) |
| Asymptotes: \(y=-\dfrac{\pi}{2}, y=\dfrac{\pi}{2}\) | Asymptotes: \(y=0, y=\pi\) |
Inverse Trigonometric Functions: Unit Circle
When we deal with inverse trigonometric functions, the unit circle is still a very helpful tool. While we typically think about using the unit circle to solve trigonometric functions, the same unit circle can be used to solve, or evaluate, the inverse trigonometric functions.
Before we get to the unit circle itself, let's take a look at another, simpler tool. The diagrams below can be used to help us remember from which quadrants the inverse trigonometric functions on the unit circle will come.
Fig. 3. A diagram that shows in which quadrants cosine, secant, and cotangent (and therefore their inverses) return values.
Just as the cosine, secant, and cotangent functions return values in Quadrants I and II (between 0 and 2π), their inverses, arc cosine, arc secant, and arc cotangent, do as well.
Fig. 4. A diagram that shows in which quadrants sine, cosecant, and tangent (and therefore their reciprocals) return values.
Just as the sine, cosecant, and tangent functions return values in Quadrants I and IV (between \(-\dfrac{\pi}{2}\) and \(\dfrac{\pi}{2}\)), their inverses, arc sine, arc cosecant, and arc tangent, do as well. Note that the values from Quadrant IV will be negative.
These diagrams assume the conventional restricted domains of the inverse functions.
There is a distinction between finding inverse trigonometric functions and solving for trigonometric functions.
Say we want to find \(\sin^{-1}\left( \dfrac{\sqrt{2}}{2} \right)\).
- Because of the restriction of the domain of inverse sine, we only want a result that lies in either Quadrant I or Quadrant IV of the unit circle.
- So, the only answer is \(\dfrac{\pi}{4}\).
Now, say we want to solve \(\sin(x)=\dfrac{\sqrt{2}}{2}\).
- There are no domain restrictions here.
- Therefore, on the interval of \((0, 2\pi)\) alone (or one loop around the unit circle), we get both \(\dfrac{\pi}{4}\) and \(\dfrac{3\pi}{4}\)as valid answers.
- And, over all real numbers, we get: \(\dfrac{\pi}{4}+2\pi k\) and \(\dfrac{3\pi}{4}+2\pi k\) as valid answers.
We might recall that we can use the Unit Circle to solve trigonometric functions of special angles: angles that have trigonometric values that we evaluate exactly.
Fig. 5. The unit circle.
When using the unit circle to evaluate inverse trigonometric functions, there are several things we need to keep in mind:
- If the answer is in Quadrant IV, it must be a negative answer (in other words, we go clockwise from the point (1, 0) instead of counterclockwise).
- For example, if we want to evaluate \(\sin^{-1}\left( -\dfrac{1}{2} \right)\) , our first instinct is to say the answer is \(330^o\) or \(\dfrac{11\pi}{6}\). However, since the answer must be between \(-\dfrac{\pi}{2}\) and \(\dfrac{\pi}{2}\) (the standard domain for inverse sine), we need to change our answer to the co-terminal angle \(-30^o\), or \(-\dfrac{\pi}{6}\).
- To use the unit circle to get the inverses for the reciprocal functions (secant, cosecant, and cotangent), we can take the reciprocal of what is in the parentheses and use the trigonometric functions.
- For example, if we want to evaluate \(\sec^{-1}(-\sqrt{2})\), we would look for \(\cos^{-1} \left( - \dfrac{1}{\sqrt{2}} \right)\) on the unit circle, which is the same as \(\cos^{-1} \left( - \dfrac{\sqrt{2}}{2} \right)\), which gives us \(\dfrac{3\pi}{4}\) or \(135^o\).
- Remember to check your work!
- Given any trigonometric function with a positive argument (assuming the conventional restricted domain), we should get an angle that is in Quadrant I \( 0 \leq \theta \leq \left( \dfrac{\pi}{2} \right) \).
- For the arcsin, arccsc, and arctan functions:
- If we are given a negative argument, our answer will be in Quadrant IV \(-\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}\).
- For the arccos, arcsec, and arccot functions:
- If we are given a negative argument, our answer will be in Quadrant II \(\dfrac{\pi}{2} \leq \theta \leq \pi\).
- For any argument that is outside the domains of the trigonometric functions for arcsin, arccsc, arccos, and arcsec, we will get no solution.
The Calculus of Inverse Trigonometric Functions
In calculus, we will be asked to find derivatives and integrals of inverse trigonometric functions. In this article, we present a brief overview of these topics.
For a more in-depth analysis, please refer to our articles on Derivatives of Inverse Trigonometric Functions and Integrals Resulting in Inverse Trigonometric Functions.
Derivatives of Inverse Trigonometric Functions
A surprising fact about the Derivatives of Inverse Trigonometric Functions is that they are algebraic functions, not trigonometric functions. The derivatives of inverse trigonometric functions are defined as:
\[\dfrac{d}{dx}\sin^{-1}(x)=\dfrac{1}{\sqrt{1-(x)^2}}\]
\[\dfrac{d}{dx}\cos^{-1}(x)=\dfrac{-1}{\sqrt{1+(x)^2}}\]
\[\dfrac{d}{dx}\tan^{-1}(x)=\dfrac{1}{1+(x)^2}\]
\[\dfrac{d}{dx}\cot^{-1}(x)=\dfrac{-1}{1+(x)^2}\]
\[\dfrac{d}{dx}\sec^{-1}(x)=\dfrac{1}{|x|\sqrt{(x)^2-1}}\]
\[\dfrac{d}{dx}\csc^{-1}(x)=\dfrac{-1}{|x|\sqrt{(x)^2-1}}\]
Integrals Resulting in Inverse Trigonometric Functions
Previously, we have developed the formulas for the derivatives of inverse trigonometric functions. These formulas are what we use to develop the Integrals Resulting in Inverse Trigonometric Functions. These integrals are defined as:
\[\int \dfrac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left( \dfrac{u}{a} \right)+C\]
\[\int \dfrac{du}{\sqrt{a^2+u^2}}=\dfrac{1}{a}\tan^{-1}\left( \dfrac{u}{a} \right)+C\]
\[\int \dfrac{du}{u \sqrt{a^2+u^2}}=\dfrac{1}{a}\sec^{-1}\left( \dfrac{u}{a} \right)+C\]
There are 6 inverse trigonometric functions, so why are there only three integrals? The reason for this is that the remaining three integrals are just negative versions of these three. In other words, the only difference between them is whether the integrand is positive or negative.
- Rather than memorizing three more formulas, if the integrand is negative, we can factor out -1 and evaluate using one of the three formulas above.
Inverse Trigonometric Integrals
Other than the integrals that result in the inverse trigonometric functions, there are integrals that involve the inverse trigonometric functions. These integrals are:
The inverse trigonometric integrals that involve arc sine.
\(\int sin^{-1} u du = sin^{-1}(u)+\sqrt{1-u^2}+C\)
\(\int u \sin^{-1}u du=\dfrac{2u^2-1}{4} \sin^{-1}(u)+\dfrac{u\sqrt{1-u^2}}{4}+C\)
\(\\int u^n sin^{-1}u du \dfrac{1}{n+1} \left[ u^{n+1} \sin^{-1}(u) - \int \dfrac{u^{n+1}du}{\sqrt{1-u^2}}, n \neq -1 \right]\)
The inverse trigonometric integrals that involve arc cosine.
\(\int cos^{-1}udu =cos^{-1}(u)-\sqrt{1-u^2}+C\)
\(\int cos^{-1} u du = \dfrac{1}{n+1}\left[ u^{n+1} \cos^{-1} (u)+ \int \dfrac{u^{n+1}du}{\sqrt{1-u^2}} \right], n \neq -1\)
The inverse trigonometric integrals that involve arc tangent.
\(\int tan^{-1}udu=tan^{-1}(u)-\dfrac{1}{2}ln(1+u^2)+C\)
\(\int u \tan^{-1} u du = \dfrac{u^2-1}{2}\tan^{-1}(u)+C\)
\(\int u^n tan^{-1} udu = \dfrac{1}{n+1}\left[ \dfrac{u^{n+1} du}{1+u^2}\right], n \neq -1\)
Solving Inverse Trigonometric Functions: Examples
When we solve, or evaluate, inverse trigonometric functions, the answer we get is an angle.
Evaluate \(\cos^{-1} \left( \dfrac{1}{2}\right) \).
Solution:
To evaluate this inverse trig function, we need to find an angle \(\theta\) such that \(\cos(\theta)=\dfrac{1}{2}\).
- While many angles of θ have this property, given the definition of \(\cos^{-1}\), we need the angle \(\theta\) that not only solves the equation, but also lies on the interval \([0, \pi]\) .
- Therefore, the solution is: \[\cos^{-1}\left( \dfrac{1}{2}\right) = \dfrac{\pi}{3}=60^o\]
What about the composition of a trigonometric function and its inverse?
Let's consider the two expressions:
\[\sin\left( sin^{-1}\left( \dfrac{\sqrt{2}}{2} \right) \right)\]
and
\[\sin^{-1}(\sin(\pi))\]
Solutions:
- The first expression simplifies as:
- \(\sin\left( sin^{-1} \left( \dfrac{\sqrt{2}}{2} \right) \right)=\sin\left( \dfrac{\pi}{4} \right)=\dfrac{\sqrt{2}}{2}\)
- The second expression simplifies as:
- \(\sin{-1}(\sin(\pi))=\sin^{-1}(0)=0\)
Let's think about the answer for the second expression in the example above.
So, what happened in this example?
- The issue here is that the inverse sine function is the inverse of the restricted sine function on the domain \( \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] \) . Therefore, for \(x\) in the interval \( \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] \), it is true that \(\sin^{-1}(\sin(x))=x\). However, for values of x outside this interval, this equation does not hold true, even though \(\sin^{-1}(\sin(x))\)is defined for all real numbers of \(x\).
Then, what about \(\sin(\sin^{-1}(y))\)? Does this expression have a similar issue?
Let's summarize these findings:
| The conditions for trigonometric functions and their inverses to cancel each other |
| \(\sin(\sin^{-1}(y)=y)\) if \(-1 \leq y \leq 1\) | \(\sin^{-1}(\sin(x))=x\) if \( -\dfrac{\pi}{2}\leq x \leq \dfrac{\pi}{2} \) |
| \(\cos(\cos^{-1}(y)=y)\) if \(-1 \leq y \leq 1\) | \(\cos^{-1}(\cos(x))=x\) if \( 0 \leq x \leq \pi \) |
| \(\tan(\tan^{-1}(y)=y)\) if \(-\infty \leq y \leq \infty\) | \(\tan^{-1}(\tan(x))=x\) if \( -\dfrac{\pi}{2}\leq x \leq \dfrac{\pi}{2} \) |
| \(\cot(\cot^{-1}(y)=y)\) if \(-\infty \leq y \leq \infty\) | \(\cot^{-1}(\cot(x))=x\) if \( 0 < x < \pi \) |
| \(\sec(\sec^{-1}(y)=y)\) if \(( -\infty, -1] \leq \cup [1, \infty)\) | \(\sec^{-1}(\sec(x))=x\) if \( 0 < x < \dfrac{\pi}{2} \cup \dfrac{\pi}{2} < x < \pi\) |
| \(\csc(\csc^{-1}(y)=y)\) if \(( -\infty, -1] \leq \cup [1, \infty)\) | \(\csc^{-1}(\csc(x))=x\) if \( -\dfrac{\pi}{2} < x < \-0 \cup 0 < x < \dfrac{\pi}{2} \) |
Evaluate the following expressions:
- \(\sin^{-1}\left( -\dfrac{\sqrt{3}}{2} \right)\)
- \( tan \left( \tan^{-1}\left( -\dfrac{1}{\sqrt{3}} \right) \right)\)
- \( cos^{-1} \left( \cos\left( \dfrac{5\pi}{4} \right) \right)\)
- \( sin^{-1} \left( \cos\left( \dfrac{2\pi}{3} \right) \right)\)
Solutions:
- To evaluate this inverse trig function, we need to find an angle \(\theta\) such that \(\sin(\theta) = - \dfrac{\sqrt{3}}{2}\) and \(-\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}\).
- The angle \( \theta= - \dfrac{\pi}{3} \) satisfies both of these conditions.
- Therefore, the solution is: \[\sin^{-1}\left( -\dfrac{\sqrt{3}}{2} \right)= -\dfrac{\pi}{3}\]
- To evaluate this inverse trig function, we first solve the “inner” function: \[tan^{-1}\left( - \dfrac{1}{\sqrt{3}} \right)\], and once we have that solution, we solve the “outer” function: \(tan(x)\) .
- \(\tan^{-1}\left( -\dfrac{1}{\sqrt{3}}\right)=-\dfrac{\pi}{6}\) → then plug \(-\dfrac{\pi}{6}\) into the “outer” function.
- \(tan\left( -\dfrac{\pi}{6}\right)=-\dfrac{1}{\sqrt{3}}\).
- Therefore: \[\tan \left( tan^{-1} \left( - \dfrac{1}{3} \right) \right)=-\dfrac{1}{\sqrt{3}}\] or, if we want to rationalize the denominator: \[\tan \left( tan^{-1} \left( - \dfrac{1}{3} \right) \right)=-\dfrac{1}{\sqrt{3}}=-\dfrac{\sqrt{3}}{3}\]
- To evaluate this inverse trig function, we first solve the “inner” function: \( \cos \left( \dfrac{5\pi}{4} \right)\) , and once we have that solution, we solve the “outer” function: \(\cos^{-1}\) .
- \(cos\left( \dfrac{5\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\) → then plug \(-\dfrac{\sqrt{2}}{2}\)into the “outer” function.
- \(\cos^{-1}\left( -\dfrac{\sqrt{2}}{2} \right)\). To evaluate this expression, we need to find an angle \(\theta\) such that \(\cos(\theta)=-\dfrac{\sqrt{2}}{2}\) and \(0 < \theta \leq \pi\).
- The angle \(\theta = \dfrac{3\pi}{4}\) satisfies both of these conditions.
- Therefore, the solution is: \[\cos^{-1}\left( cos \left( \dfrac{5\pi}{4} \right) \right)=\dfrac{3 \pi}{4}\]
- To evaluate this inverse trig function, we first solve the “inner” function: \(\cos \left( \dfrac{2 \pi}{3}\right)\) , and once we have that solution, we solve the “outer” function: \(\sin^{-1}(x)\) .
- \(\cos\left( \dfrac{2 \pi}{3} \right)= - \dfrac{1}{2}\) → then plug \(-\dfrac{1}{2}\) into the “outer” function.
- \(\sin\left( -\dfrac{1}{2} \right)\). To evaluate this expression, we need to find an angle \(\theta\) such that \(\sin(\theta)=-\dfrac{1}{2}\) and \(-\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}\).
- The angle \(\theta= -\dfrac{\pi}{6}\) satisfies both of these conditions.
- Therefore, the solution is: \[\sin^{-1}\left(\cos \left( \dfrac{2 \pi}{3} \right) \right)= -\dfrac{\pi}{6}\]
On most graphing calculators, you can directly evaluate inverse trigonometric functions for inverse sine, inverse cosine, and inverse tangent.
When it is not explicitly specified, we restrict the inverse trigonometric functions to the standard bounds specified in the section “inverse trigonometric functions in a table”. We saw this restriction in place in the first example.
However, there may be cases where we want to find an angle corresponding to a trigonometric value evaluated within a different specified bound. In such cases, it is useful to remember the trigonometric quadrants:
Fig. 6. The trigonometric quadrants and where which trig (and therefore inverse trig) functions are positive.
Given the following, find \(theta\).
\[\sin(\theta)=-0.625\]
where
\[90^o< \theta < 270^o\]
Solution:
- Using a graphing calculator, we can find that:
- \(\sin^{-1}(-0.625)=-38.68^o=-0.675rad\)
- However, based on the given range for \(\theta\), our value should lie in the 2nd or 3rd quadrant, not in the 4thquadrant, like the answer the graphing calculator gave.
- And: given that \(\sin(\theta)\) is negative, \(\theta\) has to lie in the 3rd quadrant, not in the 2nd quadrant.
- So, we know that the final answer needs to lie in the 3rd quadrant, and \(\theta\) must be between \(180\) and \(270\) degrees.
- To get the solution based on the given range, we use the identity:
- \(\sin(\theta)=\sin(180-\theta)\)
- Therefore:
- \(\sin(-38.68^o=\sin(180-(-38.68^o))=\sin(218.68^o)\)
- Thus, we have:
- \(\theta=\sin^{-1}(-0.625)=218.68^o\)
Inverse Trigonometric Functions – Key takeaways
- An inverse trigonometric function gives you an angle that corresponds to a given value of a trigonometric function.
- In general, if we know a trigonometric ratio but not the angle, we can use an inverse trigonometric function to find the angle.
- The inverse trigonometric functions must be defined on restricted domains, where they are 1-to-1 functions.
- While there is a conventional/standard domain on which the inverse trigonometric functions are defined, remember that since trigonometric functions are periodic, there are an infinite number of intervals on which they can be defined.
- The 6 main inverse trigonometric functions are:
- Inverse sine / arc sine:
- Inverse cosine / arc cosine:
- Inverse tangent / arc cotangent:
- Inverse cosecant / arc cosecant:
- Inverse secant / arc secant:
- Inverse cotangent / arc cotangent:
- To learn more about the calculus of inverse trigonometric functions, please refer to our articles on Derivatives of Inverse Trigonometric Functions and Integrals Resulting in Inverse Trigonometric Functions.
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