https://file.aiquickdraw.com/mathbot/file/1730073401876oberxhzi.jpg

To solve this expression, we need to simplify \( \csc \left( \arccos \left( \frac{9}{x} \right) \right) \).

Let's go through the steps:

1. **Define the Angle**: Let \( \theta = \arccos \left( \frac{9}{x} \right) \). This means that \( \cos(\theta) = \frac{9}{x} \).

2. **Find \( \sin(\theta) \)**: To express \( \csc(\theta) = \frac{1}{\sin(\theta)} \), we need \( \sin(\theta) \).
   - Using the Pythagorean identity, \( \sin^2(\theta) + \cos^2(\theta) = 1 \).
   - Substitute \( \cos(\theta) = \frac{9}{x} \):  
     \[
     \sin^2(\theta) + \left( \frac{9}{x} \right)^2 = 1
     \]
     \[
     \sin^2(\theta) + \frac{81}{x^2} = 1
     \]
     \[
     \sin^2(\theta) = 1 - \frac{81}{x^2}
     \]
     \[
     \sin(\theta) = \sqrt{1 - \frac{81}{x^2}}
     \]

3. **Find \( \csc(\theta) \)**: Now that \( \sin(\theta) = \sqrt{1 - \frac{81}{x^2}} \),
   \[
   \csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{\sqrt{1 - \frac{81}{x^2}}}
   \]

Thus, the equivalent algebraic expression is:
\[
\csc \left( \arccos \left( \frac{9}{x} \right) \right) = \frac{1}{\sqrt{1 - \frac{81}{x^2}}}
\]

Would you like more details or have any questions?

Here are five related questions:
1. How would this expression change if it were \( \sec(\arccos(\frac{9}{x})) \) instead of \( \csc \)?
2. What if we had \( \sin(\arccos(\frac{9}{x})) \) instead of \( \csc \)?
3. How does understanding trigonometric identities help in simplifying such expressions?
4. Why do we use the Pythagorean identity in this simplification?
5. Could we encounter similar expressions with other inverse trigonometric functions like \( \arcsin \) or \( \arctan \)?

**Tip**: When dealing with inverse trigonometric functions, try assigning a variable to the angle, then use trigonometric identities to find equivalent expressions.

Write an algebraic expression that is equivalent to csc(arccos(9/x)).

Learn how to simplify the expression csc(arccos(9/x)) into an equivalent algebraic form. This solution involves using trigonometric identities and inverse trigonometric functions, focusing on the Pythagorean identity to express the result. Suitable for high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation