Properties of the Cotangent Function

Oct 13, 2025 by magicalceylontours.com 84 views

Introduction

In mathematics, the inverse cotangent function, denoted as $\cot^{-1}x$ , is the inverse of the cotangent function. It is defined as the angle whose cotangent is a given number. The principal value of $\cot^{-1}x$ is the value of the angle in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .

Understanding the Problem

The problem asks us to find the principal value of $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ . To solve this problem, we need to understand the properties of the cotangent function and its inverse.

Properties of the Cotangent Function

The cotangent function is defined as $\cot x = \frac{\cos x}{\sin x}$ . It is periodic with a period of $\pi$ , and its range is all real numbers.

Properties of the Inverse Cotangent Function

The inverse cotangent function, $\cot^{-1}x$ , is defined as the angle whose cotangent is a given number. It is also periodic with a period of $\pi$ , but its range is restricted to the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .

Finding the Principal Value

To find the principal value of $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ , we need to find the angle whose cotangent is $-\frac{1}{\sqrt{3}}$ . We can use the fact that $\cot \frac{\pi}{6} = \frac{\sqrt{3}}{1}$ and $\cot \frac{\pi}{3} = \frac{1}{\sqrt{3}}$ .

Using the Identity $\cot \theta = -\cot \left(\pi - \theta\right)$

We can use the identity $\cot \theta = -\cot \left(\pi - \theta\right)$ to find the angle whose cotangent is $-\frac{1}{\sqrt{3}}$ . This identity states that the cotangent of an angle is equal to the negative cotangent of its supplement.

Applying the Identity

Using the identity $\cot \theta = -\cot \left(\pi - \theta\right)$ , we can write:

\cot \left(-\frac{\pi}{3}\right) = -\cot \left(\pi - \left(-\frac{\pi}{3}\right)\right)

\cot \left(-\frac{\pi}{3}\right) = -\cot \left(\frac{4\pi}{3}\right)

Finding the Principal Value

Since $\cot \frac{4\pi}{3} = -\cot \frac{\pi}{3}$ , we can write:

\cot \left(-\frac{\pi}{3}\right) = -\cot \frac{\pi}{3}

\cot \left(-\frac{\pi}{3}\right) = -\frac{1}{\sqrt{3}}

Conclusion

Therefore, the principal value of $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ is $-\frac{\pi}{3}$ .

Final Answer

The final answer is $-\frac{\pi}{3}$ .

Discussion

The principal value of $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ is $-\frac{\pi}{3}$ . This is because the cotangent function is periodic with a period of $\pi$ , and its range is all real numbers. The inverse cotangent function, $\cot^{-1}x$ , is also periodic with a period of $\pi$ , but its range is restricted to the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . Using the identity $\cot \theta = -\cot \left(\pi - \theta\right)$ , we can find the angle whose cotangent is $-\frac{1}{\sqrt{3}}$ , which is $-\frac{\pi}{3}$ .

References

[1] "Inverse Trigonometric Functions" by Wolfram MathWorld
[2] "Cotangent Function" by Wolfram MathWorld
[3] "Inverse Cotangent Function" by Wolfram MathWorld

Keywords

Inverse cotangent function
Principal value
Cotangent function
Inverse trigonometric functions
Periodic functions
Range of functions

Introduction

In our previous article, we discussed the principal value of $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ . In this article, we will answer some frequently asked questions related to this topic.

Q1: What is the principal value of $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ ?

A1: The principal value of $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ is $-\frac{\pi}{3}$ .

Q2: Why is the principal value of $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ $-\frac{\pi}{3}$ ?

A2: The principal value of $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ is $-\frac{\pi}{3}$ because the cotangent function is periodic with a period of $\pi$ , and its range is all real numbers. The inverse cotangent function, $\cot^{-1}x$ , is also periodic with a period of $\pi$ , but its range is restricted to the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . Using the identity $\cot \theta = -\cot \left(\pi - \theta\right)$ , we can find the angle whose cotangent is $-\frac{1}{\sqrt{3}}$ , which is $-\frac{\pi}{3}$ .

Q3: What is the relationship between the cotangent function and the inverse cotangent function?

A3: The cotangent function and the inverse cotangent function are related by the identity $\cot \theta = -\cot \left(\pi - \theta\right)$ . This identity states that the cotangent of an angle is equal to the negative cotangent of its supplement.

Q4: What is the range of the inverse cotangent function?

A4: The range of the inverse cotangent function, $\cot^{-1}x$ , is restricted to the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .

Q5: How do you find the principal value of $\cot^{-1}x$ ?

A5: To find the principal value of $\cot^{-1}x$ , you need to find the angle whose cotangent is a given number. You can use the identity $\cot \theta = -\cot \left(\pi - \theta\right)$ to find the angle whose cotangent is $x$ .

Q6: What is the period of the cotangent function?

A6: The period of the cotangent function is $\pi$ .

Q7: What is the period of the inverse cotangent function?

A7: The period of the inverse cotangent function, $\cot^{-1}x$ , is also $\pi$ .

Q8: How do you use the identity $\cot \theta = -\cot \left(\pi - \theta\right)$ to find the principal value of $\cot^{-1}x$ ?

A8: To use the identity $\cot \theta = -\cot \left(\pi - \theta\right)$ to find the principal value of $\cot^{-1}x$ , you need to find the angle whose cotangent is $x$ . You can then use the identity to find the principal value of $\cot^{-1}x$ .

Q9: What is the significance of the principal value of $\cot^{-1}x$ ?

A9: The principal value of $\cot^{-1}x$ is significant because it is the value of the angle whose cotangent is a given number, and it is restricted to the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .

Q10: How do you apply the identity $\cot \theta = -\cot \left(\pi - \theta\right)$ to find the principal value of $\cot^{-1}x$ ?

A10: To apply the identity $\cot \theta = -\cot \left(\pi - \theta\right)$ to find the principal value of $\cot^{-1}x$ , you need to find the angle whose cotangent is $x$ . You can then use the identity to find the principal value of $\cot^{-1}x$ .