How To Find The Exact Value Of Sec(sin⁻¹(2√6/9))

In the realm of trigonometry, finding the exact value of expressions involving inverse trigonometric functions can be a challenging yet rewarding endeavor. This article delves into the process of evaluating the expression $\sec(\sin^{-1} \frac{2\sqrt{6}}{9})$, providing a step-by-step guide to arrive at the solution. We will explore the underlying concepts, trigonometric identities, and algebraic manipulations necessary to simplify the expression and determine its exact value. Understanding inverse trigonometric functions and their relationships with standard trigonometric functions is crucial for solving such problems. The inverse sine function, denoted as $\sin^{-1}(x)$ or $\arcsin(x)$, yields the angle whose sine is $x$. Similarly, the secant function, denoted as $\sec(x)$, is the reciprocal of the cosine function, i.e., $\sec(x) = \frac{1}{\cos(x)}$. By combining these concepts, we can unravel the intricacies of the given expression and arrive at the precise numerical answer. This article aims to provide not only the solution but also a comprehensive understanding of the underlying principles, enabling readers to tackle similar trigonometric problems with confidence. We will explore the domain and range restrictions of inverse trigonometric functions and how they influence the solution process. Additionally, we will emphasize the importance of visualizing trigonometric relationships using right-angled triangles, which can greatly simplify the problem-solving approach. By the end of this article, readers will have a solid grasp of how to evaluate expressions involving inverse trigonometric functions and will be well-equipped to tackle a wide range of trigonometric challenges.

Understanding Inverse Trigonometric Functions

To effectively solve the given expression, a solid understanding of inverse trigonometric functions is essential. Inverse trigonometric functions, also known as arcus functions, are the inverses of the standard trigonometric functions: sine, cosine, tangent, cotangent, secant, and cosecant. The inverse sine function, denoted as $\sin^{-1}(x)$ or $\arcsin(x)$, returns the angle whose sine is $x$. It is crucial to remember that the domain of $\sin^{-1}(x)$ is $[-1, 1]$, and its range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. This restricted range is necessary to ensure that the inverse sine function is a well-defined function. Similarly, the inverse cosine function, $\cos^{-1}(x)$ or $\arccos(x)$, yields the angle whose cosine is $x$. Its domain is also $[-1, 1]$, but its range is $[0, \pi]$. The inverse tangent function, $\tan^{-1}(x)$ or $\arctan(x)$, gives the angle whose tangent is $x$. Its domain is all real numbers, and its range is $(-\frac{\pi}{2}, \frac{\pi}{2})$. The other inverse trigonometric functions, such as $\cot^{-1}(x)$, $\sec^{-1}(x)$, and $\csc^{-1}(x)$, can be expressed in terms of the inverse sine, cosine, and tangent functions. For instance, $\sec^{-1}(x) = \cos^{-1}(\frac{1}{x})$. When dealing with expressions involving inverse trigonometric functions, it is often helpful to visualize a right-angled triangle. By letting the inverse trigonometric function represent an angle in the triangle, we can use the trigonometric ratios (sine, cosine, tangent) to determine the sides of the triangle. This geometric approach can simplify the process of evaluating complex trigonometric expressions. Understanding the domain and range restrictions of inverse trigonometric functions is crucial for avoiding errors in calculations. For example, when evaluating $\sin^{-1}(\frac{2\sqrt{6}}{9})$, we need to ensure that $\frac{2\sqrt{6}}{9}$ lies within the domain of the inverse sine function, which is $[-1, 1]$. In this case, $\frac{2\sqrt{6}}{9}$ is approximately 0.544, which falls within the domain. By carefully considering the properties of inverse trigonometric functions, we can navigate trigonometric problems with greater accuracy and efficiency.

Step-by-Step Solution

Let's now proceed with solving the expression $\sec(\sin^{-1} \frac{2\sqrt{6}}{9})$.

1. Let $\theta = \sin^{-1} \frac{2\sqrt{6}}{9}$.

   • This substitution simplifies the expression to $\sec(\theta)$.

2. From the definition of the inverse sine function, we have $\sin(\theta) = \frac{2\sqrt{6}}{9}$.

   • Since $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$, we can visualize a right-angled triangle where the opposite side is $2\sqrt{6}$ and the hypotenuse is 9.

3. Use the Pythagorean theorem to find the adjacent side. Let the adjacent side be $a$. Then,

   • $a^2 + (2\sqrt{6})^2 = 9^2$
   • $a^2 + 4 \cdot 6 = 81$
   • $a^2 + 24 = 81$
   • $a^2 = 57$
   • $a = \sqrt{57}$

4. Now we have a right-angled triangle with opposite side $2\sqrt{6}$, adjacent side $\sqrt{57}$, and hypotenuse 9.

5. Find $\cos(\theta)$. Since $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$, we have

   • $\cos(\theta) = \frac{\sqrt{57}}{9}$

6. Find $\sec(\theta)$. Since $\sec(\theta) = \frac{1}{\cos(\theta)}$, we have

   • $\sec(\theta) = \frac{1}{\frac{\sqrt{57}}{9}} = \frac{9}{\sqrt{57}}$

7. Rationalize the denominator:

   • $\sec(\theta) = \frac{9}{\sqrt{57}} \cdot \frac{\sqrt{57}}{\sqrt{57}} = \frac{9\sqrt{57}}{57}$

8. Simplify the fraction:

   • $\sec(\theta) = \frac{9\sqrt{57}}{57} = \frac{3\sqrt{57}}{19}$

Therefore, the exact value of the expression $\sec(\sin^{-1} \frac{2\sqrt{6}}{9})$ is $\frac{3\sqrt{57}}{19}$.

Key Trigonometric Identities Used

Throughout the solution process, several key trigonometric identities were employed. These identities are fundamental to simplifying and evaluating trigonometric expressions. Understanding and memorizing these identities is crucial for success in trigonometry. The primary identity used in this solution is the Pythagorean theorem, which relates the sides of a right-angled triangle. Specifically, the Pythagorean theorem states that in a right-angled triangle with sides $a$ and $b$ and hypotenuse $c$, the following relationship holds: $a^2 + b^2 = c^2$. This theorem allowed us to determine the length of the adjacent side of the triangle, given the lengths of the opposite side and the hypotenuse. Another essential identity is the definition of the sine, cosine, and secant functions in terms of the sides of a right-angled triangle. The sine of an angle is defined as the ratio of the opposite side to the hypotenuse, i.e., $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$. The cosine of an angle is defined as the ratio of the adjacent side to the hypotenuse, i.e., $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$. The secant of an angle is the reciprocal of the cosine, i.e., $\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{\text{hypotenuse}}{\text{adjacent}}$. These definitions were crucial in relating the given inverse sine function to the sides of the right-angled triangle and subsequently finding the value of the secant function. In addition to these basic identities, the reciprocal identity $\sec(\theta) = \frac{1}{\cos(\theta)}$ played a key role in transforming the expression from cosine to secant. By mastering these trigonometric identities, one can approach a wide range of trigonometric problems with greater confidence and efficiency. These identities provide the building blocks for simplifying complex expressions and establishing relationships between different trigonometric functions.

Common Mistakes to Avoid

When evaluating expressions involving inverse trigonometric functions, several common mistakes can lead to incorrect solutions. Being aware of these pitfalls can help avoid errors and ensure accurate results. One common mistake is neglecting the domain and range restrictions of inverse trigonometric functions. As mentioned earlier, the inverse sine function, $\sin^{-1}(x)$, has a domain of $[-1, 1]$ and a range of $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Similarly, the inverse cosine function, $\cos^{-1}(x)$, has a domain of $[-1, 1]$ and a range of $[0, \pi]$. Failing to consider these restrictions can lead to incorrect angles and ultimately the wrong answer. For example, if one attempts to evaluate $\sin^{-1}(2)$, the result will be undefined because 2 is outside the domain of the inverse sine function. Another frequent error is incorrectly applying trigonometric identities. It is crucial to remember the correct definitions and relationships between trigonometric functions and their inverses. For instance, confusing the definitions of sine, cosine, and secant in terms of the sides of a right-angled triangle can lead to errors in calculating the ratios. Another mistake is failing to rationalize the denominator when expressing the final answer. While $\frac{9}{\sqrt{57}}$ is mathematically equivalent to $\frac{3\sqrt{57}}{19}$, the latter is considered the simplified form and is generally preferred. Rationalizing the denominator ensures that the answer is presented in its most concise and conventional form. Additionally, errors can arise from algebraic manipulations, such as incorrectly applying the Pythagorean theorem or simplifying fractions. Careful attention to detail and a systematic approach can help prevent these mistakes. It is always a good practice to double-check each step of the solution to ensure accuracy. By being mindful of these common mistakes and taking the necessary precautions, one can improve their ability to solve trigonometric problems involving inverse trigonometric functions accurately and efficiently.

Conclusion

In conclusion, finding the exact value of the expression $\sec(\sin^{-1} \frac{2\sqrt{6}}{9})$ requires a solid understanding of inverse trigonometric functions, trigonometric identities, and algebraic manipulations. By systematically applying these concepts, we arrived at the solution $\frac{3\sqrt{57}}{19}$. This process involved visualizing a right-angled triangle, using the Pythagorean theorem to find the missing side, and applying the definitions of sine, cosine, and secant. Furthermore, we emphasized the importance of considering the domain and range restrictions of inverse trigonometric functions and avoiding common mistakes. Trigonometry is a fascinating and powerful branch of mathematics with numerous applications in various fields, including physics, engineering, and computer science. Mastering trigonometric concepts and problem-solving techniques is essential for anyone pursuing studies or careers in these areas. By understanding the fundamental principles and practicing problem-solving, one can develop a strong foundation in trigonometry and confidently tackle complex trigonometric challenges. This article has provided a comprehensive guide to evaluating expressions involving inverse trigonometric functions, and we hope it has been helpful in enhancing your understanding of trigonometry. Remember, practice is key to success in mathematics, so continue to explore trigonometric problems and refine your skills. With dedication and perseverance, you can unlock the beauty and power of trigonometry and its applications.

Final Answer: A. $\sec(\sin^{-1} \frac{2\sqrt{6}}{9}) = \frac{3\sqrt{57}}{19}$