If `cosalpha+cosbeta=a ,sinalpha+sinbeta=b and alpha-beta=2theta,` then `(cos3theta)/(costheta)=`

Options

A. `a^(2) + b^(2) -2`
B. `a^(2) +b^(2) -3`
C. `3-a^(2) - 3`
D. ` (a^(2) +b^(2))/(4)`

Answer

To solve the problem, we need to find the value of \(\frac{\cos 3\theta}{\cos \theta}\) given that \(\cos \alpha + \cos \beta = a\), \(\sin \alpha + \sin \beta = b\), and \(\alpha - \beta = 2\theta\).

### Step-by-Step Solution:

1. **Square the equations for cosine and sine:**
\[ (\cos \alpha + \cos \beta)^2 = a^2 \quad \text{(1)} \]
\[ (\sin \alpha + \sin \beta)^2 = b^2 \quad \text{(2)} \]

2. **Expand both equations:**
From equation (1):
\[ \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta = a^2 \]
From equation (2):
\[ \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta = b^2 \]

3. **Add the two equations:**
\[ (\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) = a^2 + b^2 \]
Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ 1 + 1 + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) = a^2 + b^2 \]
\[ 2 + 2\cos(\alpha - \beta) = a^2 + b^2 \]

4. **Substitute \(\alpha - \beta = 2\theta\):**
\[ 2 + 2\cos(2\theta) = a^2 + b^2 \]
Dividing through by 2:
\[ 1 + \cos(2\theta) = \frac{a^2 + b^2}{2} \]
Rearranging gives:
\[ \cos(2\theta) = \frac{a^2 + b^2}{2} - 1 \]

5. **Use the half-angle formula for \(\cos(2\theta)\):**
The half-angle formula states:
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \]
Setting the two expressions for \(\cos(2\theta)\) equal:
\[ 2\cos^2(\theta) - 1 = \frac{a^2 + b^2}{2} - 1 \]
Simplifying gives:
\[ 2\cos^2(\theta) = \frac{a^2 + b^2}{2} \]
\[ \cos^2(\theta) = \frac{a^2 + b^2}{4} \quad \text{(3)} \]

6. **Find \(\cos(3\theta)\) using the formula:**
The formula for \(\cos(3\theta)\) is:
\[ \cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta) \]

7. **Substitute \(\cos^2(\theta)\) into the formula:**
First, find \(\cos^3(\theta)\):
\[ \cos^3(\theta) = \cos(\theta) \cdot \cos^2(\theta) = \cos(\theta) \cdot \frac{a^2 + b^2}{4} \]
Now substitute back into the equation for \(\cos(3\theta)\):
\[ \cos(3\theta) = 4\left(\cos(\theta) \cdot \frac{a^2 + b^2}{4}\right) - 3\cos(\theta) \]
\[ = (a^2 + b^2)\cos(\theta) - 3\cos(\theta) \]
\[ = (a^2 + b^2 - 3)\cos(\theta) \]

8. **Finally, calculate \(\frac{\cos(3\theta)}{\cos(\theta)}\):**
\[ \frac{\cos(3\theta)}{\cos(\theta)} = a^2 + b^2 - 3 \]

### Final Answer:
\[ \frac{\cos(3\theta)}{\cos(\theta)} = a^2 + b^2 - 3 \]