`( cos alpha - cos beta )^(2) + ( sin alpha - sin beta )^(2) = `

Options

A. `4 cos^(2) ( alpha + beta )//2`
B. ` 4 sin^(2) ( alpha + beta )//2`
C. ` 4 sin^(2) ( alpha -beta )//2`
D. `4 cos^(2) ( alpha - beta ) //2`

Answer

To solve the expression \( ( \cos \alpha - \cos \beta )^{2} + ( \sin \alpha - \sin \beta )^{2} \), we can follow these steps:

### Step 1: Expand the squares
We start by expanding both squares in the expression:

\[ ( \cos \alpha - \cos \beta )^{2} = \cos^{2} \alpha - 2 \cos \alpha \cos \beta + \cos^{2} \beta \]

\[ ( \sin \alpha - \sin \beta )^{2} = \sin^{2} \alpha - 2 \sin \alpha \sin \beta + \sin^{2} \beta \]

### Step 2: Combine the expanded terms
Now, we combine the results from Step 1:

\[ ( \cos \alpha - \cos \beta )^{2} + ( \sin \alpha - \sin \beta )^{2} = (\cos^{2} \alpha + \sin^{2} \alpha) + (\cos^{2} \beta + \sin^{2} \beta) - 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) \]

### Step 3: Use the Pythagorean identity
Using the Pythagorean identity \( \cos^{2} \theta + \sin^{2} \theta = 1 \):

\[ \cos^{2} \alpha + \sin^{2} \alpha = 1 \]
\[ \cos^{2} \beta + \sin^{2} \beta = 1 \]

So, we can substitute these into our expression:

\[ 1 + 1 - 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) \]

### Step 4: Simplify the expression
Now, we simplify the expression:

\[ 2 - 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) \]

### Step 5: Use the cosine of the difference identity
Recall the cosine of the difference identity:

\[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \]

Substituting this into our expression gives:

\[ 2 - 2 \cos(\alpha - \beta) \]

### Step 6: Factor out the common term
Factor out the common term:

\[ 2(1 - \cos(\alpha - \beta)) \]

### Step 7: Use the identity for \(1 - \cos \theta\)
Using the identity \(1 - \cos \theta = 2 \sin^{2} \left( \frac{\theta}{2} \right)\):

\[ 1 - \cos(\alpha - \beta) = 2 \sin^{2} \left( \frac{\alpha - \beta}{2} \right) \]

Substituting this back into our expression:

\[ 2 \cdot 2 \sin^{2} \left( \frac{\alpha - \beta}{2} \right) = 4 \sin^{2} \left( \frac{\alpha - \beta}{2} \right) \]

### Final Result
Thus, we have:

\[ ( \cos \alpha - \cos \beta )^{2} + ( \sin \alpha - \sin \beta )^{2} = 4 \sin^{2} \left( \frac{\alpha - \beta}{2} \right) \]