Determine If The Following Statement Is True Or False: \tan^{-1} X = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right ]

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Introduction

In mathematics, the inverse trigonometric functions are used to find the angle whose trigonometric function is a given value. The inverse tangent function, denoted by $\tan^{-1} x$, is used to find the angle whose tangent is $x$. Similarly, the inverse sine function, denoted by $\sin^{-1} x$, is used to find the angle whose sine is $x$. In this article, we will discuss the statement $\tan^{-1} x = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$ and determine if it is true or false.

Understanding the Inverse Trigonometric Functions

Before we proceed to analyze the given statement, let's understand the inverse trigonometric functions. The inverse tangent function, $\tan^{-1} x$, is defined as the angle $\theta$ such that $\tan \theta = x$. Similarly, the inverse sine function, $\sin^{-1} x$, is defined as the angle $\theta$ such that $\sin \theta = x$.

Analyzing the Given Statement

The given statement is $\tan^{-1} x = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$. To analyze this statement, we need to find the expression for $\tan^{-1} x$ and compare it with the expression for $\sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$.

Expression for $\tan^{-1} x$

The expression for $\tan^{-1} x$ can be found using the following formula:

$$\tan^{-1} x = \arctan x$$

This formula states that the inverse tangent of $x$ is equal to the arctangent of $x$.

Expression for $\sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$

To find the expression for $\sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$, we can use the following formula:

$$\sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right) = \arcsin \left( \frac{x}{\sqrt{1+x^2}} \right)$$

This formula states that the inverse sine of $\frac{x}{\sqrt{1+x^2}}$ is equal to the arcsine of $\frac{x}{\sqrt{1+x^2}}$.

Comparing the Expressions

Now that we have the expressions for $\tan^{-1} x$ and $\sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$, we can compare them to determine if the given statement is true or false.

Using the formula for $\tan^{-1} x$, we can write:

$$\tan^{-1} x = \arctan x$$

Using the formula for $\sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$, we can write:

$$\sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right) = \arcsin \left( \frac{x}{\sqrt{1+x^2}} \right)$$

Now, let's compare the two expressions:

$$\arctan x = \arcsin \left( \frac{x}{\sqrt{1+x^2}} \right)$$

Conclusion

Based on the comparison of the expressions, we can conclude that the given statement $\tan^{-1} x = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$ is TRUE.

The expression for $\tan^{-1} x$ is equal to the expression for $\sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$, which means that the two inverse trigonometric functions are equal.

Proof

To prove that the given statement is true, we can use the following steps:

1. Start with the expression for $\tan^{-1} x$:

$$\tan^{-1} x = \arctan x$$

2. Use the trigonometric identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$ to rewrite the expression:

$$\arctan x = \arctan \left( \frac{\sin \theta}{\cos \theta} \right)$$

3. Use the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ to rewrite the expression:

$$\arctan \left( \frac{\sin \theta}{\cos \theta} \right) = \arctan \left( \frac{\sin \theta}{\sqrt{1-\sin^2 \theta}} \right)$$

4. Use the trigonometric identity $\sin \theta = \frac{x}{\sqrt{1+x^2}}$ to rewrite the expression:

$$\arctan \left( \frac{\sin \theta}{\sqrt{1-\sin^2 \theta}} \right) = \arctan \left( \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1-\left( \frac{x}{\sqrt{1+x^2}} \right)^2}} \right)$$

5. Simplify the expression:

$$\arctan \left( \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1-\left( \frac{x}{\sqrt{1+x^2}} \right)^2}} \right) = \arctan \left( \frac{x}{\sqrt{1+x^2}} \right)$$

6. Use the definition of the inverse sine function to rewrite the expression:

$$\arctan \left( \frac{x}{\sqrt{1+x^2}} \right) = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$$

7. Therefore, we have:

$$\tan^{-1} x = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$$

This proves that the given statement is true.

Conclusion

In conclusion, the given statement $\tan^{-1} x = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$ is true. The expression for $\tan^{-1} x$ is equal to the expression for $\sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$, which means that the two inverse trigonometric functions are equal.

Introduction

In our previous article, we discussed the relationship between the inverse tangent function and the inverse sine function. We proved that the statement $\tan^{-1} x = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$ is true. In this article, we will answer some frequently asked questions about the relationship between inverse trigonometric functions.

Q: What is the relationship between the inverse tangent function and the inverse sine function?

A: The inverse tangent function and the inverse sine function are related through the following equation:

$$\tan^{-1} x = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$$

This equation shows that the inverse tangent function and the inverse sine function are equal for all values of $x$.

Q: How do I prove that the inverse tangent function and the inverse sine function are equal?

A: To prove that the inverse tangent function and the inverse sine function are equal, you can use the following steps:

1. Start with the expression for $\tan^{-1} x$:

$$\tan^{-1} x = \arctan x$$

2. Use the trigonometric identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$ to rewrite the expression:

$$\arctan x = \arctan \left( \frac{\sin \theta}{\cos \theta} \right)$$

3. Use the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ to rewrite the expression:

$$\arctan \left( \frac{\sin \theta}{\cos \theta} \right) = \arctan \left( \frac{\sin \theta}{\sqrt{1-\sin^2 \theta}} \right)$$

4. Use the trigonometric identity $\sin \theta = \frac{x}{\sqrt{1+x^2}}$ to rewrite the expression:

$$\arctan \left( \frac{\sin \theta}{\sqrt{1-\sin^2 \theta}} \right) = \arctan \left( \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1-\left( \frac{x}{\sqrt{1+x^2}} \right)^2}} \right)$$

5. Simplify the expression:

$$\arctan \left( \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1-\left( \frac{x}{\sqrt{1+x^2}} \right)^2}} \right) = \arctan \left( \frac{x}{\sqrt{1+x^2}} \right)$$

6. Use the definition of the inverse sine function to rewrite the expression:

$$\arctan \left( \frac{x}{\sqrt{1+x^2}} \right) = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$$

7. Therefore, we have:

$$\tan^{-1} x = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$$

This proves that the inverse tangent function and the inverse sine function are equal.

Q: What are some common applications of inverse trigonometric functions?

A: Inverse trigonometric functions have many applications in mathematics, physics, and engineering. Some common applications include:

• Trigonometry: Inverse trigonometric functions are used to find the angles of triangles.
• Physics: Inverse trigonometric functions are used to find the angles of motion and the positions of objects.
• Engineering: Inverse trigonometric functions are used to design and analyze mechanical systems, electrical circuits, and other engineering systems.
• Computer Science: Inverse trigonometric functions are used in computer graphics, game development, and other areas of computer science.

Q: How do I use inverse trigonometric functions in real-world applications?

A: To use inverse trigonometric functions in real-world applications, you can follow these steps:

1. Identify the problem: Identify the problem you are trying to solve and determine which inverse trigonometric function is needed.
2. Choose the correct function: Choose the correct inverse trigonometric function based on the problem.
3. Apply the function: Apply the inverse trigonometric function to the given values.
4. Solve the problem: Solve the problem using the result of the inverse trigonometric function.

For example, if you are designing a mechanical system and need to find the angle of a triangle, you can use the inverse tangent function to find the angle.

Conclusion

In conclusion, inverse trigonometric functions are an important part of mathematics and have many applications in real-world problems. By understanding the relationship between inverse trigonometric functions and how to use them in real-world applications, you can solve a wide range of problems and make informed decisions.