If `cos 25^0 + sin 25^0 = K`, then `cos 50^0` is equal to

Options

A. `sqrt(2-p^(2))`
B. `-sqrt(2-p^(2))`
C. `psqrt(2-p^(2))`
D. `-psqrt(2-p^(2))`

Explanation

To solve the problem, we need to find the value of \( \cos 50^\circ \) given that \( \cos 25^\circ + \sin 25^\circ = K \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \cos 25^\circ + \sin 25^\circ = K \]

2. **Square both sides:**
\[ (\cos 25^\circ + \sin 25^\circ)^2 = K^2 \]

3. **Expand the left-hand side using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \):**
\[ \cos^2 25^\circ + \sin^2 25^\circ + 2 \sin 25^\circ \cos 25^\circ = K^2 \]

4. **Use the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \):**
\[ 1 + 2 \sin 25^\circ \cos 25^\circ = K^2 \]

5. **Recognize that \( 2 \sin x \cos x = \sin 2x \):**
\[ 1 + \sin 50^\circ = K^2 \]

6. **Rearranging gives us:**
\[ \sin 50^\circ = K^2 - 1 \]

7. **Now, we want to find \( \cos 50^\circ \). Use the identity \( \sin^2 x + \cos^2 x = 1 \):**
\[ \cos^2 50^\circ = 1 - \sin^2 50^\circ \]

8. **Substituting \( \sin 50^\circ \) into the equation:**
\[ \cos^2 50^\circ = 1 - (K^2 - 1)^2 \]

9. **Expanding \( (K^2 - 1)^2 \):**
\[ (K^2 - 1)^2 = K^4 - 2K^2 + 1 \]
So,
\[ \cos^2 50^\circ = 1 - (K^4 - 2K^2 + 1) \]

10. **Simplifying gives:**
\[ \cos^2 50^\circ = 1 - K^4 + 2K^2 - 1 = 2K^2 - K^4 \]

11. **Taking the square root to find \( \cos 50^\circ \):**
\[ \cos 50^\circ = \sqrt{2K^2 - K^4} \]

### Final Answer:
\[ \cos 50^\circ = \sqrt{2K^2 - K^4} \]