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To simplify the expression \(\cos(\arcsin(x))\), we can approach it geometrically using a right triangle.

### Steps to Simplify:

1. **Interpret the Inverse Sine Function**:
   \[
   y = \arcsin(x)
   \]
   This means that \(\sin(y) = x\). In terms of a right triangle, this implies that the opposite side of the angle \(y\) is \(x\), and the hypotenuse is 1 (since \(\sin(y) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{1}\)).

2. **Find the Adjacent Side**:
   To find \(\cos(y)\), we need the adjacent side. Using the Pythagorean theorem, the adjacent side is:
   \[
   \text{adjacent} = \sqrt{1^2 - x^2} = \sqrt{1 - x^2}
   \]

3. **Find \(\cos(y)\)**:
   Since \(\cos(y)\) is the ratio of the adjacent side to the hypotenuse:
   \[
   \cos(\arcsin(x)) = \cos(y) = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}
   \]

### Final Answer:
\[
\cos(\arcsin(x)) = \sqrt{1 - x^2}
\]

Let me know if you'd like further clarification or if you have any questions!

Here are five related questions to deepen your understanding:
1. How do we define the domain of the arcsin function?
2. What is the range of values for \(\cos(\arcsin(x))\) when \(x\) is in the domain of \(\arcsin(x)\)?
3. Can the identity \(\cos(\arcsin(x)) = \sqrt{1 - x^2}\) be extended to complex numbers?
4. How would you express \(\sin(\arccos(x))\)?
5. What happens if you apply both \(\cos\) and \(\arcsin\) to a number outside the typical range of \(\arcsin\)?

**Tip:** When working with inverse trigonometric functions, it's helpful to think in terms of right triangles to simplify expressions involving compositions of these functions.

Learn how to simplify the trigonometric expression cos(arcsin x) by applying the Pythagorean Theorem and inverse trigonometric functions. This problem is suitable for high school students (Grades 9-12) and covers key concepts in trigonometry and right triangle geometry.

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