Evaluate the expression. sin ⁡ − 1 ( cos ⁡ 2 π 3 ) \sin^{-1}\left(\cos\frac{2\pi}{3}\right) sin − 1 ( cos 3 2 π )

− π 6 -\frac{\pi}{6} − 6 π

Evaluate the expression. tan ⁡ − 1 ( tan ⁡ 2 π 3 ) \tan^{-1}\left(\tan\frac{2\pi}{3}\right) tan − 1 ( tan 3 2 π )

− π 3 -\frac{\pi}{3} − 3 π

0. Functions

Inverse Trigonometric Functions

0. Functions

Inverse Trigonometric Functions: 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 within defined intervals. For instance, the inverse cosine function yields angles between 0 and π, while the inverse sine and tangent functions yield angles between -π/2 and π/2. Evaluating composite functions involves solving the inner function first, considering the unit circle and the specified intervals to ensure valid outputs. Understanding these principles is crucial for accurately navigating trigonometric expressions and their applications in various mathematical contexts.

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. While a standard cosine function takes an angle and returns a cosine value, the inverse cosine function starts with 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 instance, 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, only one of these angles is valid 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 $ \frac{5\pi}{3} $ is not a valid solution since it falls outside the specified range. Therefore, the only valid angle for $ \cos^{-1} \left( \frac{1}{2} \right) $ is $ \frac{\pi}{3} $. In summary, when evaluating inverse cosine expressions, it is essential to find the corresponding angle on the unit circle while adhering to the defined interval of $ 0 $ to $ \pi $.

example

Example 1

Video duration:

Example 1 Video Summary

To evaluate the expression for the inverse cosine of $-\frac{\sqrt{3}}{2}$, we need to determine the angle whose cosine value equals $-\frac{\sqrt{3}}{2}$. When working 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 from $0$ to $\pi$ (or $0$ to $180^\circ$).

In this case, since we are dealing with a negative cosine value, we can conclude that the angle must be located in the second quadrant, where cosine values are negative. The cosine of $\frac{\sqrt{3}}{2}$ corresponds to angles in the first quadrant, specifically at $\frac{\pi}{6}$ (or $30^\circ$). To find the equivalent angle in the second quadrant, we can use the reference angle:

For the angle in the second quadrant, we can express it as:

\[\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\]

Thus, the angle for which the cosine equals $-\frac{\sqrt{3}}{2}$ is $\frac{5\pi}{6}$. This is our final answer.

Problem

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

$0$

$π$

$pi over 2$

$3 pi over 2$

Problem

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

$0$

$pi over 2$

$π$

$3 pi over 2$

Problem

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

$pi over 4$

$3 pi over 4$

$5 pi over 4$

$7 pi over 4$

concept

Inverse Sine

Video duration:

Inverse Sine Video Summary

Understanding the inverse sine function is essential for evaluating expressions involving angles and their sine values. The inverse sine function, denoted as sin^-1(x) or arcsin(x), is defined only for values of x in the interval of [-1, 1]. The output, or the angle θ, is restricted to the interval of [-π/2, π/2]. This means that when you input a value into the inverse sine function, you will receive an angle that lies within this specified range.

To visualize the inverse sine function, we can reflect the sine graph over the line y = x. However, since the sine function is not one-to-one over its entire domain, we only consider the portion of the sine graph from -π/2 to π/2 for our reflection. This results in the graph of the inverse sine function, which is crucial for understanding how to evaluate expressions.

For example, to evaluate sin^-1(1/2), we need to find 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. Since π/6 lies within the interval of [-π/2, π/2], it is the correct solution.

In another example, consider sin^-1(-√2/2). We seek an angle θ where sin(θ) = -√2/2. On the unit circle, this occurs at -π/4 in the fourth quadrant. Although 7π/4 also yields a sine value of -√2/2, it is outside the specified interval for the inverse sine function. Therefore, the solution is -π/4, which is within the acceptable range.

In summary, when evaluating inverse sine expressions, always ensure that the input value is between -1 and 1, and that the resulting angle is within -π/2 to π/2. This approach will help you accurately determine the angles corresponding to given sine values.

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 within the range of -π/2 to π/2. This means that the output of the inverse sine function will always yield angles in the first and fourth quadrants.

In the first quadrant (from 0 to π/2), all sine values are positive, while in the fourth quadrant (from -π/2 to 0), sine values are negative. Since we are looking for the inverse sine of a negative value, specifically -1, we can conclude that the solution must lie in the fourth quadrant.

Now, we need to identify the specific angle in the fourth quadrant where the sine value equals -1. The angle that satisfies this condition is -π/2, where the sine function reaches its minimum value. Therefore, the solution to the expression sin^-1(-1) is:

-π/2

This result indicates that the angle corresponding to the sine of negative one is -π/2.

Problem

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

$0$

$pi over 2$

$π$

$- \frac{π}{2}$

Problem

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

$pi over 6$

$3 pi over 2$

$pi over 3$

$- \frac{π}{3}$

Problem

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

$pi over 4$

$- \frac{π}{4}$

$3 pi over 4$

$7 pi over 4$

concept

Inverse Tangent

Video duration:

Inverse Tangent Video Summary

When working with inverse trigonometric functions, it's essential to understand the specific intervals for each function. For the inverse tangent function, the input values can range from negative infinity to 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, which is necessary for it to have an inverse.

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 of the graph between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ for our reflection. This results in the graph of the inverse tangent function, which is defined within the specified interval.

For example, to evaluate $\tan^{-1}(\sqrt{3})$, we seek an angle $\theta$ such that $\tan(\theta) = \sqrt{3}$. Within the interval from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, the angle that satisfies this condition is $\frac{\pi}{3}$, since $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$.

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

When evaluating inverse trigonometric functions using a calculator, the process is straightforward. You typically press the second function key followed by the corresponding trigonometric function key (e.g., for inverse sine, press 2nd and then sine). Ensure your calculator is set to radian mode unless otherwise specified, as this is the standard for trigonometric calculations.

By understanding these concepts and practicing evaluations, you can confidently work with inverse trigonometric functions in various mathematical contexts.

example

Example 3

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Example 3 Video Summary

To evaluate the expression for the inverse tangent of negative square root of 3, we start by interpreting it as finding the angle whose tangent is equal to $-\sqrt{3}$. The range for the inverse tangent function, or arctan, is limited to the interval $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. This range corresponds to the first and fourth quadrants of the unit circle.

In the first quadrant, tangent values are positive, while in the fourth quadrant, they are negative. Since we are dealing with the inverse tangent of a negative value, we focus on the fourth quadrant. The reference angle for $\sqrt{3}$ is $\frac{\pi}{3}$, which means that the angle in the fourth quadrant that has a tangent of $-\sqrt{3}$ is $-\frac{\pi}{3}$.

Thus, the solution to the expression is:

\[\arctan(-\sqrt{3}) = -\frac{\pi}{3}\]

This result indicates that the inverse tangent of negative square root of 3 is $-\frac{\pi}{3}$, confirming our understanding of the relationship between angles and their tangent values within the specified range.

Problem

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

$0$

$pi over 2$

$π$

$- \frac{π}{2}$

Problem

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

$0$

$pi over 4$

$pi over 2$

$- \frac{π}{4}$

Problem

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

$pi over 6$

$pi over 3$

$- \frac{π}{6}$

$- \frac{π}{3}$

Problem

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

$negative 5$

$negative 3.38$

$1.37$

$78.69$

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

concept

Evaluate Composite Functions - Values on Unit Circle

Video duration:

Evaluate Composite Functions - Values on Unit Circle Video Summary

Understanding composite trigonometric functions is essential for solving complex expressions involving inverse trigonometric functions. When tasked with finding the sine of the inverse cosine of a value, such as $ \frac{1}{2} $, it is helpful to break the problem down into manageable parts. This approach allows us to evaluate the inner function first before addressing the outer function.

To find the sine of the inverse cosine of $ \frac{1}{2} $, we start by determining the angle whose cosine is $ \frac{1}{2} $. Referring to the unit circle, we find that $ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} $. Thus, the inverse cosine of $ \frac{1}{2} $ is $ \frac{\pi}{3} $. Next, we evaluate the outer function, which is the sine of $ \frac{\pi}{3} $. Again, using the unit circle, we see that $ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} $. Therefore, the sine of the inverse cosine of $ \frac{1}{2} $ is $ \frac{\sqrt{3}}{2} $.

In another example, we can explore the cosine of the inverse tangent of $ 0 $. Here, we first find the angle whose tangent is $ 0 $. The tangent function equals $ 0 $ at $ 0 $ and $ \pi $, but since the range of the inverse tangent function is limited to $ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) $, we select $ 0 $ as our angle. Thus, the inverse tangent of $ 0 $ is $ 0 $. Now, we find the cosine of $ 0 $, which is $ 1 $. Therefore, the cosine of the inverse tangent of $ 0 $ is $ 1 $.

Lastly, consider the expression for the inverse cosine of the sine of $ \frac{\pi}{3} $. We start by evaluating the sine of $ \frac{\pi}{3} $, which we previously determined to be $ \frac{\sqrt{3}}{2} $. Now, we need to find the angle whose cosine is $ \frac{\sqrt{3}}{2} $. The cosine function equals $ \frac{\sqrt{3}}{2} $ at $ \frac{\pi}{6} $, which is within the range of the inverse cosine function, $ [0, \pi] $. Thus, the inverse cosine of the sine of $ \frac{\pi}{3} $ is $ \frac{\pi}{6} $.

By systematically evaluating composite trigonometric functions, we can simplify complex expressions and arrive at accurate solutions. This method not only enhances our understanding of trigonometric identities but also strengthens our problem-solving skills in trigonometry.

Problem

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

$0$

$1$

$-1$

$\frac12$

Problem

Evaluate the expression.
$\sin^{-1}\left(\cos\frac{2\pi}{3}\right)$

$\frac{\pi}{6}$

$\frac{5\pi}{6}$

$\frac{\pi}{3}$

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

example

Example 1

Video duration:

Example 1 Video Summary

To evaluate the expression for the secant of the inverse cosine of $-\frac{\sqrt{3}}{2}$, we start by determining the angle whose cosine equals $-\frac{\sqrt{3}}{2}$. This requires us to consider the range of the inverse cosine function, which is limited to angles between $0$ and $\pi$.

From the unit circle, we find that the angle $ \frac{5\pi}{6} $ has a cosine value of $-\frac{\sqrt{3}}{2}$. Thus, we can express the original problem as finding the secant of $ \frac{5\pi}{6} $.

Recall that the secant function is defined as the reciprocal of the cosine function. Therefore, we can rewrite the secant of $ \frac{5\pi}{6} $ as:

\[\sec\left(\frac{5\pi}{6}\right) = \frac{1}{\cos\left(\frac{5\pi}{6}\right)}\]

Since we already established that $\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$, we substitute this value into our equation:

\[\sec\left(\frac{5\pi}{6}\right) = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}}\]

To simplify this expression, we rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$:

\[-\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}\]

Thus, the final result for the secant of the inverse cosine of $-\frac{\sqrt{3}}{2}$ is:

\[-\frac{2\sqrt{3}}{3}\]

This concludes the evaluation, confirming that the secant of the inverse cosine of $-\frac{\sqrt{3}}{2}$ is $-\frac{2\sqrt{3}}{3}$.

concept

Evaluate Composite Functions - Special Cases

Video duration:

Evaluate Composite Functions - Special Cases Video Summary

When working with composite trigonometric functions, it's essential to understand the relationship between trigonometric functions and their inverses. For example, consider the expression $\cos^{-1}(\cos(\frac{11\pi}{6}))$ . While it may seem intuitive to cancel the cosine and its inverse, this approach is incorrect due to the defined intervals of inverse functions.

To solve $\cos^{-1}(\cos(\frac{11\pi}{6}))$ , start by evaluating the inner function, which is $\cos(\frac{11\pi}{6})$ . Referring to the unit circle, we find that $\cos(\frac{11\pi}{6}) = \frac{\sqrt{3}}{2}$ . Next, we evaluate the outer function, $\cos^{-1}(\frac{\sqrt{3}}{2})$ . The inverse cosine function is defined for angles between $0$ and $\pi$ . Within this interval, the angle whose cosine is $\frac{\sqrt{3}}{2}$ is $\frac{\pi}{6}$ . Thus, the final answer is $\cos^{-1}(\cos(\frac{11\pi}{6})) = \frac{\pi}{6}$ , which is not equal to $\frac{11\pi}{6}$ .

In another example, consider $\sin(\sin^{-1}(2))$ . Here, the inner function is $\sin^{-1}(2)$ . The inverse sine function is only defined for values between $-1$ and $1$ , and since $2$ is outside this range, $\sin^{-1}(2)$ is undefined. Therefore, attempting to evaluate $\sin(\sin^{-1}(2))$ leads to an error, highlighting the importance of considering the defined intervals of inverse trigonometric functions.

In summary, when dealing with composite trigonometric functions, always evaluate from the inside out and be mindful of the intervals for inverse functions to avoid incorrect assumptions and errors.

example

Example 2

Video duration:

Example 2 Video Summary

To evaluate the expression of the inverse sine of the sine of $\frac{\pi}{6}$, it is essential to understand the properties of inverse trigonometric functions and their defined intervals. Initially, we calculate the sine of $\frac{\pi}{6}$, which corresponds to a value of $\frac{1}{2}$ based on the unit circle.

Next, we need to find the inverse sine of $\frac{1}{2}$. The inverse sine function, denoted as $\sin^{-1}(x)$ or $\arcsin(x)$, is defined only for angles within the interval $[- \frac{\pi}{2}, \frac{\pi}{2}]$. Therefore, we seek an angle within this range for which the sine equals $\frac{1}{2}$.

Within the specified interval, the angle that satisfies this condition is $\frac{\pi}{6}$, since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$. Thus, we conclude that:

\[\sin^{-1}\left(\sin\left(\frac{\pi}{6}\right)\right) = \frac{\pi}{6}\]

It is important to note that while it may seem permissible to cancel the inverse sine with the sine directly, this is only valid when the original angle lies within the defined interval of the inverse function. In this case, since $\frac{\pi}{6}$ is indeed within the interval $[- \frac{\pi}{2}, \frac{\pi}{2}]$, we can safely conclude that the final answer is $\frac{\pi}{6}$.

Always remember to verify the interval when working with composite trigonometric functions to ensure accurate results.

Problem

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

$\frac{\pi}{3}$

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

$\pi$

Undefined

Problem

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

$0$

$\frac{\pi}{2}$

$\pi$

Undefined

Problem

Evaluate the expression.
$\tan^{-1}\left(\tan\frac{2\pi}{3}\right)$

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

$\frac{\pi}{3}$

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

$\frac{5\pi}{3}$

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More sets

Inverse Trigonometric Functions

0. Functions

3 problems

Topic

0. Functions - Part 1 of 2

7 topics 8 problems

Chapter

0. Functions - Part 2 of 2

7 topics 9 problems

Chapter

Here’s what students ask on this topic:

What is the inverse cosine of 1/2?

The inverse cosine of 1/2, denoted as cos^-1(1/2), is the angle whose cosine value is 1/2. Using the unit circle, we find that the angle is π/3. Therefore, cos^-1(1/2) = π/3.

How do you evaluate the inverse sine of 1/2?

To evaluate the inverse sine of 1/2, denoted as sin^-1(1/2), we need to find the angle whose sine value is 1/2. Referring to the unit circle, we see that this angle is π/6. Therefore, sin^-1(1/2) = π/6.

What is the inverse tangent of √3?

The inverse tangent of √3, denoted as tan^-1(√3), is the angle whose tangent value is √3. From the unit circle, we find that this angle is π/3. Therefore, tan^-1(√3) = π/3.

How do you find the sine of the inverse cosine of 1/2?

To find the sine of the inverse cosine of 1/2, we first determine the inverse cosine of 1/2, which is π/3. Then, we find the sine of π/3. Using the unit circle, we see that sin(π/3) = √3/2. Therefore, the sine of the inverse cosine of 1/2 is √3/2.

What is the inverse cosine of the cosine of 11π/6?

To find the inverse cosine of the cosine of 11π/6, we first determine the cosine of 11π/6, which is √3/2. Then, we find the inverse cosine of √3/2. Considering the interval for inverse cosine (0 to π), the angle is π/6. Therefore, the inverse cosine of the cosine of 11π/6 is π/6.

Your Calculus tutors

Inverse Trigonometric Functions: 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)

Evaluate Composite Functions - Values on Unit Circle

Evaluate Composite Functions - Values on Unit Circle Video Summary

Evaluate the expression.cos(sin−11)\cos\left(\sin^{-1}1\right)cos(sin−11)

Evaluate the expression.sin−1(cos2π3)\sin^{-1}\left(\cos\frac{2\pi}{3}\right)sin−1(cos32π)

Example 1

Example 1 Video Summary

Evaluate Composite Functions - Special Cases

Evaluate Composite Functions - Special Cases Video Summary

Example 2

Example 2 Video Summary

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

Evaluate the expression.cos−1(cos(π2))\cos^{-1}\left(\cos\left(\frac{\pi}{2}\right)\right)cos−1(cos(2π))

Evaluate the expression.tan−1(tan2π3)\tan^{-1}\left(\tan\frac{2\pi}{3}\right)tan−1(tan32π)

Do you want more practice?

Here’s what students ask on this topic:

What is the inverse cosine of 1/2?

How do you evaluate the inverse sine of 1/2?

What is the inverse tangent of √3?

How do you find the sine of the inverse cosine of 1/2?

What is the inverse cosine of the cosine of 11π/6?

Your Calculus 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.
$\cos^{-1}\left(\frac14\right)$

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

Evaluate the expression.
$\sin^{-1}\left(\cos\frac{2\pi}{3}\right)$

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

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

Evaluate the expression.
$\tan^{-1}\left(\tan\frac{2\pi}{3}\right)$