https://file.aiquickdraw.com/mathbot/file/1724239066069kn9qfwfd.jpg

Given the image, let's solve the problem step by step.

### Problem Breakdown:

1. **Given:**
   - Triangle \(DEF\) is a right triangle with \(\angle DFE = 90^\circ\).
   - \(DF = 4\), \(FE = 3\).
   - The measure of \(\angle DFG = x^\circ\) and \(\angle GFE = y^\circ\).
   - We need to find the value of \(\cos x^\circ - \sin y^\circ\).

2. **Step 1: Find \(\cos x\) and \(\sin y\):**

   - First, calculate the hypotenuse \(DE\) of triangle \(DEF\) using the Pythagorean theorem:
     \[
     DE = \sqrt{DF^2 + FE^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
     \]
   - Now, since \(\cos x = \cos(\angle DFG)\), and \(\angle DFG = \angle DFE\), we have:
     \[
     \cos x = \cos(\angle DFE) = \frac{FE}{DE} = \frac{3}{5}
     \]
   - Similarly, since \(\sin y = \sin(\angle GFE)\), and \(\angle GFE = \angle DFE\), we have:
     \[
     \sin y = \sin(\angle DFE) = \frac{DF}{DE} = \frac{4}{5}
     \]

3. **Step 2: Calculate \(\cos x - \sin y\):**

   \[
   \cos x - \sin y = \frac{3}{5} - \frac{4}{5} = \frac{3 - 4}{5} = \frac{-1}{5} = -\frac{1}{5}
   \]

### Final Answer:
\[
\cos x^\circ - \sin y^\circ = -\frac{1}{5}
\]

Would you like more details or have any other questions?

Here are some related questions you might find interesting:

1. How would the answer change if the triangle were not a right triangle?
2. What is the significance of the trigonometric identities used here?
3. Can the Pythagorean theorem be applied in non-right triangles?
4. How would you find the angles \(x\) and \(y\) explicitly?
5. How does the relationship between the sides and angles of a triangle affect trigonometric values?

**Tip:** Remember that in a right triangle, the Pythagorean theorem is key to connecting side lengths and trigonometric ratios!

Explore how to solve a trigonometric problem involving a right triangle DEF with step-by-step instructions. Learn about the Pythagorean theorem and trigonometric ratios used to find cosine and sine values. Suitable for high school students studying trigonometry.

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