Math Problem Statement

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

Solution

Step 1: Understanding the function f(x)=sin[cos1(x+25)]f(x) = \sin \left[ \cos^{-1} \left( x + \frac{2}{5} \right) \right]

This function contains two operations:

  • cos1(x+25)\cos^{-1} \left( x + \frac{2}{5} \right), the inverse cosine (arc cosine), which gives an angle θ\theta whose cosine is x+25x + \frac{2}{5}.
  • sin(θ)\sin \left( \theta \right), the sine of that angle.

Step 2: Simplifying the expression

Let’s set: θ=cos1(x+25)\theta = \cos^{-1} \left( x + \frac{2}{5} \right) By the definition of inverse cosine, we know: cos(θ)=x+25\cos(\theta) = x + \frac{2}{5} Now, using the Pythagorean identity for sine: sin(θ)=1cos2(θ)\sin(\theta) = \sqrt{1 - \cos^2(\theta)} Substitute cos(θ)=x+25\cos(\theta) = x + \frac{2}{5}: sin(θ)=1(x+25)2\sin(\theta) = \sqrt{1 - \left( x + \frac{2}{5} \right)^2} Thus, the simplified form of f(x)f(x) is: f(x)=1(x+25)2f(x) = \sqrt{1 - \left( x + \frac{2}{5} \right)^2}

Step 3: Finding the domain of f(x)f(x)

For the square root to be real, the argument inside the square root must be non-negative: 1(x+25)201 - \left( x + \frac{2}{5} \right)^2 \geq 0 This simplifies to: (x+25)21\left( x + \frac{2}{5} \right)^2 \leq 1 Taking the square root of both sides: 1x+251-1 \leq x + \frac{2}{5} \leq 1 Now, subtract 25\frac{2}{5} from all parts of the inequality: 125x125-1 - \frac{2}{5} \leq x \leq 1 - \frac{2}{5} Simplifying further: 75x35-\frac{7}{5} \leq x \leq \frac{3}{5} Thus, the domain of f(x)f(x) is: [75,35]\boxed{ \left[ -\frac{7}{5}, \frac{3}{5} \right] }

Final Answer:

The simplified form of f(x)f(x) is: f(x)=1(x+25)2f(x) = \sqrt{1 - \left( x + \frac{2}{5} \right)^2} The domain of f(x)f(x) is: [75,35]\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(cos1(x))\sin \left( \cos^{-1}(x) \right)?
  2. How does the inverse cosine function restrict the range of inputs?
  3. Can you compute f(0)f(0) for this function?
  4. How would the domain change if 25\frac{2}{5} were a different constant?
  5. What are the properties of the arc-cosine function?

Tip: The inverse trigonometric functions, like cos1(x)\cos^{-1}(x), always return angles in specific ranges, typically restricted by the properties of the corresponding trigonometric function.

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Math Problem Analysis

Mathematical Concepts

Inverse Trigonometric Functions
Pythagorean Identity
Domain and Range of Functions

Formulas

f(x) = sin[cos^(-1)(x + 2/5)]
sin(θ) = √(1 - cos^2(θ))

Theorems

Pythagorean Identity
Properties of Inverse Cosine Function

Suitable Grade Level

Grades 10-12