Evaluate the expression. sin ⁡ − 1 1 \sin^{-1}1 sin − 1 1

− π 2 -\frac{\pi}{2} − 2 π

Evaluate the expression. sin ⁡ − 1 ( 3 2 ) \sin^{-1}\left(\frac{\sqrt3}{2}\right) sin − 1 ( 2 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> )

− π 3 -\frac{\pi}{3} − 3 π

Evaluate the expression. sin ⁡ − 1 ( 2 2 ) \sin^{-1}\left(\frac{\sqrt2}{2}\right) sin − 1 ( 2 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> )

− π 4 -\frac{\pi}{4} − 4 π

Evaluate the expression. tan ⁡ − 1 0 \tan^{-1}0 tan − 1 0

− π 2 -\frac{\pi}{2} − 2 π

Evaluate the expression. tan ⁡ − 1 1 \tan^{-1}1 tan − 1 1

− π 4 -\frac{\pi}{4} − 4 π

Evaluate the expression. tan ⁡ − 1 ( − 3 3 ) \tan^{-1}\left(-\frac{\sqrt3}{3}\right) tan − 1 ( − 3 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> )

− π 6 -\frac{\pi}{6} − 6 π

− π 3 -\frac{\pi}{3} − 3 π

11. Inverse Trigonometric Functions and Basic Trig Equations

Inverse Sine, Cosine, & Tangent

11. Inverse Trigonometric Functions and Basic Trig Equations

Inverse Sine, Cosine, & Tangent: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Inverse trigonometric functions, such as inverse sine, cosine, and tangent, allow us to find angles corresponding to specific values. For inverse cosine, the output angle is restricted to the interval [0, π], while inverse sine outputs angles within [-π/2, π/2]. Inverse tangent can take any real number input but outputs angles in the same interval as inverse sine. Understanding these intervals is crucial for evaluating expressions accurately using the unit circle, ensuring solutions fall within the defined ranges.

concept

Inverse Cosine

Video duration:

Inverse Cosine Video Summary

Inverse trigonometric functions, such as the inverse cosine, serve to reverse the action of their corresponding trigonometric functions. For instance, while the cosine function takes an angle and produces a cosine value, the inverse cosine function takes a cosine value and returns the angle. This relationship is crucial when evaluating expressions like the inverse cosine of $ \frac{1}{2} $.

To find the angle corresponding to a cosine value, we can utilize the unit circle. For example, when evaluating $ \cos^{-1} \left( \frac{1}{2} \right) $, we seek the angle $ \theta $ such that $ \cos(\theta) = \frac{1}{2} $. On the unit circle, we identify that $ \theta = \frac{\pi}{3} $ in the first quadrant has a cosine value of $ \frac{1}{2} $. Additionally, $ \theta = \frac{5\pi}{3} $ in the fourth quadrant also yields a cosine value of $ \frac{1}{2} $. However, not all angles are valid solutions for the inverse cosine function.

To determine the correct angle, we must consider the defined range of the inverse cosine function, which is limited to angles between $ 0 $ and $ \pi $. This restriction ensures that the inverse cosine function remains a valid function, as the cosine function is not one-to-one over its entire range. By reflecting only the segment of the cosine graph from $ 0 $ to $ \pi $ over the line $ y = x $, we obtain the graph of the inverse cosine function.

Given this interval, we can conclude that while $ \frac{5\pi}{3} $ is a valid angle for the cosine function, it does not fall within the specified range for the inverse cosine. Therefore, the only valid solution for $ \cos^{-1} \left( \frac{1}{2} \right) $ is $ \frac{\pi}{3} $.

In summary, when evaluating inverse cosine expressions, it is essential to find angles that correspond to the given cosine values, ensuring that the solutions lie within the interval $ [0, \pi] $. This approach allows for accurate evaluations of inverse trigonometric functions using the unit circle.

example

Example 1

Video duration:

Example 1 Video Summary

To evaluate the expression for the inverse cosine of $-\frac{\sqrt{3}}{2}$, we can reframe the problem as finding the angle whose cosine value equals $-\frac{\sqrt{3}}{2}$. When dealing with inverse trigonometric functions, it's essential to remember the range of the angles. For the inverse cosine function, the angles are restricted to the interval $[0, \pi]$.

Within this interval, cosine values are positive in the first quadrant (from $0$ to $\frac{\pi}{2}$) and negative in the second quadrant (from $\frac{\pi}{2}$ to $\pi$). Since we are looking for a negative cosine value, we can conclude that the angle must be in the second quadrant.

Next, we identify the specific angle where the cosine equals $-\frac{\sqrt{3}}{2}$. The angle that satisfies this condition is $\frac{5\pi}{6}$, as the cosine of $\frac{5\pi}{6}$ indeed equals $-\frac{\sqrt{3}}{2}$. Therefore, the solution to the expression is:

$ \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6} $

Problem

Evaluate the expression.
$\cos^{-1}\left(-1\right)$

$0$

$\pi$

$\frac{\pi}{2}$

$\frac{3\pi}{2}$

Problem

Evaluate the expression.
$\cos^{-1}\left(0\right)$

$0$

$\frac{\pi}{2}$

$\pi$

$\frac{3\pi}{2}$

Problem

Evaluate the expression.
$\cos^{-1}\left(-\frac{\sqrt2}{2}\right)$

$\frac{\pi}{4}$

$\frac{3\pi}{4}$

$\frac{5\pi}{4}$

$\frac{7\pi}{4}$

concept

Inverse Sine

Video duration:

Inverse Sine Video Summary

Understanding the inverse sine function is essential for evaluating expressions involving trigonometric functions. The inverse sine function, denoted as sin^-1(x) or arcsin(x), is derived from the sine function by reflecting its graph over the line y = x. However, since the sine function is not one-to-one over its entire range, we restrict our focus to the interval from -π/2 to π/2 for the inverse sine function. This ensures that each input value corresponds to a unique output angle.

When evaluating inverse sine expressions, the input values must lie within the range of -1 to 1. The output angles, which are the results of the inverse sine function, will always fall within the specified interval of -π/2 to π/2. For example, to evaluate sin^-1(1/2), we seek an angle θ such that sin(θ) = 1/2. Referring to the unit circle, we find that θ = π/6 is the angle that satisfies this condition, as it lies within the required interval.

In another example, consider sin^-1(-√2/2). Here, we look for an angle θ where sin(θ) = -√2/2. The angle -π/4 meets this criterion, as it is located in the fourth quadrant and falls within the specified interval. Although 7π/4 also yields a sine value of -√2/2, it is outside the acceptable range for the inverse sine function.

In summary, when working with the inverse sine function, always ensure that the input values are between -1 and 1, and that the resulting angles are confined to the interval from -π/2 to π/2. This approach will facilitate accurate evaluations of inverse sine expressions.

example

Example 2

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Example 2 Video Summary

To evaluate the expression for the inverse sine of negative one, we need to determine the angle whose sine value equals negative one. The inverse sine function, denoted as sin^-1(x), is defined for angles between -\frac{\pi}{2} and \frac{\pi}{2}. This range corresponds to the first and fourth quadrants of the unit circle.

In the first quadrant (from 0 to \frac{\pi}{2}), all sine values are positive. Conversely, in the fourth quadrant (from -\frac{\pi}{2} to 0), sine values are negative. Since we are looking for the inverse sine of a negative value, our solution must lie in the fourth quadrant.

Specifically, we need to find the angle where the sine equals negative one. This occurs at -\frac{\pi}{2}, where the sine function reaches its minimum value. Therefore, the solution to the expression sin^-1(-1) is:

-\frac{\pi}{2}

This result indicates that the angle corresponding to the sine of negative one is -90 degrees or -\frac{\pi}{2} radians.

Problem

Evaluate the expression.
$\sin^{-1}1$

$0$

$\frac{\pi}{2}$

$\pi$

$-\frac{\pi}{2}$

Problem

Evaluate the expression.
$\sin^{-1}\left(\frac{\sqrt3}{2}\right)$

$\frac{\pi}{6}$

$\frac{3\pi}{2}$

$\frac{\pi}{3}$

$-\frac{\pi}{3}$

Problem

Evaluate the expression.
$\sin^{-1}\left(\frac{\sqrt2}{2}\right)$

$\frac{\pi}{4}$

$-\frac{\pi}{4}$

$\frac{3\pi}{4}$

$\frac{7\pi}{4}$

concept

Inverse Tangent

Video duration:

Inverse Tangent Video Summary

When working with inverse trigonometric functions, it's essential to understand the specific intervals for which these functions are defined. For the inverse tangent function, the input values can range from negative infinity to positive infinity, but the output angles are restricted to the interval between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. This restriction ensures that the function remains one-to-one, allowing us to find unique angle values.

To visualize the inverse tangent function, we can reflect the tangent function's graph over the line $y = x$. However, since the tangent function is not one-to-one over its entire range, we only consider the segment between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ for this reflection.

For example, to evaluate $\tan^{-1}(\sqrt{3})$, we seek an angle $\theta$ such that $\tan(\theta) = \sqrt{3}$. Within the specified interval, we find that $\theta = \frac{\pi}{3}$ satisfies this condition, as the tangent of $\frac{\pi}{3}$ equals $\sqrt{3}$.

Similarly, to find $\tan^{-1}(-1)$, we look for an angle $\theta$ where $\tan(\theta) = -1$. In the interval from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, this occurs at $\theta = -\frac{\pi}{4}$.

When evaluating inverse trigonometric functions using a calculator, the process is straightforward. You typically press the 'second' button followed by the corresponding trigonometric function (sine, cosine, or tangent) and then input the value you wish to evaluate. It's important to ensure that your calculator is set to radian mode unless specified otherwise.

By understanding these concepts and practicing various examples, you can confidently evaluate inverse trigonometric functions and apply them in different mathematical contexts.

example

Example 3

Video duration:

Example 3 Video Summary

To evaluate the expression for the inverse tangent of negative square root of 3, we need to determine the angle whose tangent value equals negative square root of 3. The inverse tangent function, denoted as tan^-1(x), is defined within the interval of (-π/2, π/2). This range is crucial as it restricts the possible angles we can consider.

When analyzing the unit circle, we recognize that tangent values are positive in the first quadrant and negative in the fourth quadrant. Since we are dealing with the inverse tangent of a negative value, our solution must lie in the fourth quadrant.

In the fourth quadrant, the reference angle corresponding to π/3 is -π/3. The tangent of this angle is negative square root of 3, which confirms that:

tan(-π/3) = -√3

Thus, the solution to the expression tan^-1(-√3) is:

-π/3

This result indicates that the inverse tangent of negative square root of 3 is -π/3, providing a clear understanding of the relationship between the angle and its tangent value.

Problem

Evaluate the expression.
$\tan^{-1}0$

$0$

$\frac{\pi}{2}$

$\pi$

$-\frac{\pi}{2}$

Problem

Evaluate the expression.
$\tan^{-1}1$

$0$

$\frac{\pi}{4}$

$\frac{\pi}{2}$

$-\frac{\pi}{4}$

Problem

Evaluate the expression.
$\tan^{-1}\left(-\frac{\sqrt3}{3}\right)$

$\frac{\pi}{6}$

$\frac{\pi}{3}$

$-\frac{\pi}{6}$

$-\frac{\pi}{3}$

Problem

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.
$\tan^{-1}\left(5\right)$

– 5

– 3.38

1.37

78.69

Problem

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.
$\sin^{-1}\left(-\frac13\right)$

0.34

– 0.34

– 0.33

– 19.47

Problem

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.
$\cos^{-1}\left(\frac14\right)$

75.52

1.82

1.32

0.97

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11. Inverse Trigonometric Functions and Basic Trig Equations

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11. Inverse Trigonometric Functions and Basic Trig Equations

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Inverse Sine, Cosine, & Tangent definitions
11. Inverse Trigonometric Functions and Basic Trig Equations
15 Terms

Here’s what students ask on this topic:

What is the inverse cosine of 1/2?

The inverse cosine of 1/2 is the angle whose cosine value is 1/2. Using the unit circle, we find that cos(π/3) = 1/2. Since the range of the inverse cosine function is [0, π], the correct answer is π/3.

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How do you evaluate the inverse sine of -√2/2?

To evaluate the inverse sine of -√2/2, we need to find the angle whose sine value is -√2/2. On the unit circle, sin(-π/4) = -√2/2. Since the range of the inverse sine function is [-π/2, π/2], the correct answer is -π/4.

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What is the range of the inverse tangent function?

The range of the inverse tangent function is the set of angles between -π/2 and π/2. This means that the output of the inverse tangent function will always be an angle within this interval.

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How do you find the inverse tangent of √3?

To find the inverse tangent of √3, we need to determine the angle whose tangent value is √3. On the unit circle, tan(π/3) = √3. Since the range of the inverse tangent function is [-π/2, π/2], the correct answer is π/3.

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What is the inverse sine of 1/2?

The inverse sine of 1/2 is the angle whose sine value is 1/2. Using the unit circle, we find that sin(π/6) = 1/2. Since the range of the inverse sine function is [-π/2, π/2], the correct answer is π/6.

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How do you evaluate the inverse cosine of -1?

To evaluate the inverse cosine of -1, we need to find the angle whose cosine value is -1. On the unit circle, cos(π) = -1. Since the range of the inverse cosine function is [0, π], the correct answer is π.

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Inverse Sine, Cosine, & Tangent: Videos & Practice Problems

Inverse Cosine

Inverse Cosine Video Summary

Example 1

Example 1 Video Summary

Evaluate the expression.cos−1(−1)\cos^{-1}\left(-1\right)cos−1(−1)

Evaluate the expression.cos−1(0)\cos^{-1}\left(0\right)cos−1(0)

Evaluate the expression.cos−1(−22)\cos^{-1}\left(-\frac{\sqrt2}{2}\right)cos−1(−22)

Inverse Sine

Inverse Sine Video Summary

Example 2

Example 2 Video Summary

Evaluate the expression.sin−11\sin^{-1}1sin−11

Evaluate the expression.sin−1(32)\sin^{-1}\left(\frac{\sqrt3}{2}\right)sin−1(23)

Evaluate the expression.sin−1(22)\sin^{-1}\left(\frac{\sqrt2}{2}\right)sin−1(22)

Inverse Tangent

Inverse Tangent Video Summary

Example 3

Example 3 Video Summary

Evaluate the expression.tan−10\tan^{-1}0tan−10

Evaluate the expression.tan−11\tan^{-1}1tan−11

Evaluate the expression.tan−1(−33)\tan^{-1}\left(-\frac{\sqrt3}{3}\right)tan−1(−33)

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.tan−1(5)\tan^{-1}\left(5\right)tan−1(5)

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.sin−1(−13)\sin^{-1}\left(-\frac13\right)sin−1(−31)

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.cos−1(14)\cos^{-1}\left(\frac14\right)cos−1(41)

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Go over this topic definitions with flashcards

Here’s what students ask on this topic:

What is the inverse cosine of 1/2?

How do you evaluate the inverse sine of -√2/2?

What is the range of the inverse tangent function?

How do you find the inverse tangent of √3?

What is the inverse sine of 1/2?

How do you evaluate the inverse cosine of -1?

Your Precalculus tutors

Evaluate the expression.
$\cos^{-1}\left(-1\right)$

Evaluate the expression.
$\cos^{-1}\left(0\right)$

Evaluate the expression.
$\cos^{-1}\left(-\frac{\sqrt2}{2}\right)$

Evaluate the expression.
$\sin^{-1}1$

Evaluate the expression.
$\sin^{-1}\left(\frac{\sqrt3}{2}\right)$

Evaluate the expression.
$\sin^{-1}\left(\frac{\sqrt2}{2}\right)$

Evaluate the expression.
$\tan^{-1}0$

Evaluate the expression.
$\tan^{-1}1$

Evaluate the expression.
$\tan^{-1}\left(-\frac{\sqrt3}{3}\right)$

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.
$\tan^{-1}\left(5\right)$

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.
$\sin^{-1}\left(-\frac13\right)$

Evaluate the expression using a calculator. Express your answer in radians, rounding to two decimal places.
$\cos^{-1}\left(\frac14\right)$