If `cos theta+cos phi=a` and `sin theta - sin phi=b`, then: `2cos(theta + phi)`=

Options

A. `a^(2)-b^(2)-2`
B. `a^(2)-b^(2)+2`
C. `a^(2)+ b^(2)=2`
D. `a^(2) + b^(2)-2`

Answer

To solve the problem, we need to find the expression for \( 2 \cos(\theta + \phi) \) given the equations \( \cos \theta + \cos \phi = a \) and \( \sin \theta - \sin \phi = b \).

### Step-by-Step Solution:

1. **Start with the given equations:**
\[ \cos \theta + \cos \phi = a \quad \text{(1)} \]
\[ \sin \theta - \sin \phi = b \quad \text{(2)} \]

2. **Square both equations:**
- For equation (1):
\[ (\cos \theta + \cos \phi)^2 = a^2 \]
Expanding this, we get:
\[ \cos^2 \theta + 2 \cos \theta \cos \phi + \cos^2 \phi = a^2 \quad \text{(3)} \]

- For equation (2):
\[ (\sin \theta - \sin \phi)^2 = b^2 \]
Expanding this, we get:
\[ \sin^2 \theta - 2 \sin \theta \sin \phi + \sin^2 \phi = b^2 \quad \text{(4)} \]

3. **Add equations (3) and (4):**
\[ (\cos^2 \theta + \sin^2 \theta) + (\cos^2 \phi + \sin^2 \phi) + 2 \cos \theta \cos \phi - 2 \sin \theta \sin \phi = a^2 + b^2 \]
Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \) and \( \cos^2 \phi + \sin^2 \phi = 1 \):
\[ 1 + 1 + 2 \cos \theta \cos \phi - 2 \sin \theta \sin \phi = a^2 + b^2 \]
Simplifying gives:
\[ 2 + 2 (\cos \theta \cos \phi - \sin \theta \sin \phi) = a^2 + b^2 \]

4. **Recognize the cosine addition formula:**
The expression \( \cos \theta \cos \phi - \sin \theta \sin \phi \) can be rewritten using the cosine of the sum:
\[ \cos(\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi \]
Thus, we have:
\[ 2 + 2 \cos(\theta + \phi) = a^2 + b^2 \]

5. **Isolate \( 2 \cos(\theta + \phi) \):**
\[ 2 \cos(\theta + \phi) = a^2 + b^2 - 2 \]

6. **Final result:**
Therefore, we can conclude that:
\[ 2 \cos(\theta + \phi) = a^2 + b^2 - 2 \]