The simplified form of \(f(x)=\sin [\cos ^{-1}(x+\frac{2}{5})]\). and find the domain of f(x)

### Step 1: Understanding the function \( f(x) = \sin \left[ \cos^{-1} \left( x + \frac{2}{5} \right) \right] \)

This function contains two operations:
- \( \cos^{-1} \left( x + \frac{2}{5} \right) \), the inverse cosine (arc cosine), which gives an angle \( \theta \) whose cosine is \( x + \frac{2}{5} \).
- \( \sin \left( \theta \right) \), the sine of that angle.

### Step 2: Simplifying the expression
Let’s set:
\[
 \theta = \cos^{-1} \left( x + \frac{2}{5} \right)
\]
By the definition of inverse cosine, we know:
\[
 \cos(\theta) = x + \frac{2}{5}
\]
Now, using the Pythagorean identity for sine:
\[
 \sin(\theta) = \sqrt{1 - \cos^2(\theta)}
\]
Substitute \( \cos(\theta) = x + \frac{2}{5} \):
\[
 \sin(\theta) = \sqrt{1 - \left( x + \frac{2}{5} \right)^2}
\]
Thus, the simplified form of \( f(x) \) is:
\[
 f(x) = \sqrt{1 - \left( x + \frac{2}{5} \right)^2}
\]

### Step 3: Finding the domain of \( f(x) \)
For the square root to be real, the argument inside the square root must be non-negative:
\[
 1 - \left( x + \frac{2}{5} \right)^2 \geq 0
\]
This simplifies to:
\[
 \left( x + \frac{2}{5} \right)^2 \leq 1
\]
Taking the square root of both sides:
\[
 -1 \leq x + \frac{2}{5} \leq 1
\]
Now, subtract \( \frac{2}{5} \) from all parts of the inequality:
\[
 -1 - \frac{2}{5} \leq x \leq 1 - \frac{2}{5}
\]
Simplifying further:
\[
 -\frac{7}{5} \leq x \leq \frac{3}{5}
\]
Thus, the domain of \( f(x) \) is:
\[
 \boxed{ \left[ -\frac{7}{5}, \frac{3}{5} \right] }
\]

### Final Answer:
The simplified form of \( f(x) \) is:
\[
 f(x) = \sqrt{1 - \left( x + \frac{2}{5} \right)^2}
\]
The domain of \( f(x) \) is:
\[
 \boxed{ \left[ -\frac{7}{5}, \frac{3}{5} \right] }
\]

Would you like more details or have any questions?

Here are some related questions to explore:
1. What is the geometric interpretation of \( \sin \left( \cos^{-1}(x) \right) \)?
2. How does the inverse cosine function restrict the range of inputs?
3. Can you compute \( f(0) \) for this function?
4. How would the domain change if \( \frac{2}{5} \) were a different constant?
5. What are the properties of the arc-cosine function?

**Tip:** The inverse trigonometric functions, like \( \cos^{-1}(x) \), always return angles in specific ranges, typically restricted by the properties of the corresponding trigonometric function.

The simplified form of f(x)=sin[cos^(-1)(x + 2/5)] and find the domain of f(x).

Explore the detailed solution to simplify f(x) = sin[cos^(-1)(x + 2/5)] and find its domain. Learn how inverse trigonometric functions and Pythagorean identity play key roles in this problem, ideal for high school students in Grades 10-12.

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