If `A = cos ^(2) alpha + cos ^2(alpha + beta)- 2 cos alpha cos beta cos (alpha + beta),` then

Options

A. `A gt 1`
B. `0 le A le 1`
C. `-1 le A lt 0`
D. `A lt -1`

Answer

To solve the problem, we need to simplify the expression given for \( A \):

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

### Step 1: Use the cosine addition formula
Recall the cosine addition formula:
\[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]
Using this, we can express \( \cos^2(\alpha + \beta) \):
\[ \cos^2(\alpha + \beta) = (\cos \alpha \cos \beta - \sin \alpha \sin \beta)^2 \]

### Step 2: Expand the square
Now, expand \( \cos^2(\alpha + \beta) \):
\[ \cos^2(\alpha + \beta) = \cos^2 \alpha \cos^2 \beta - 2 \cos \alpha \cos \beta \sin \alpha \sin \beta + \sin^2 \alpha \sin^2 \beta \]

### Step 3: Substitute back into the expression for \( A \)
Substituting this back into the expression for \( A \):
\[ A = \cos^2 \alpha + \left( \cos^2 \alpha \cos^2 \beta - 2 \cos \alpha \cos \beta \sin \alpha \sin \beta + \sin^2 \alpha \sin^2 \beta \right) - 2 \cos \alpha \cos \beta \left( \cos \alpha \cos \beta - \sin \alpha \sin \beta \right) \]

### Step 4: Simplify the expression
Now, simplify the expression:
\[ A = \cos^2 \alpha + \cos^2 \alpha \cos^2 \beta - 2 \cos \alpha \cos \beta \sin \alpha \sin \beta + \sin^2 \alpha \sin^2 \beta - 2 \cos^2 \alpha \cos^2 \beta + 2 \cos \alpha \cos \beta \sin \alpha \sin \beta \]
This simplifies to:
\[ A = \cos^2 \alpha - \cos^2 \alpha \cos^2 \beta + \sin^2 \alpha \sin^2 \beta \]

### Step 5: Factor out common terms
Factoring out \( \cos^2 \alpha \):
\[ A = \cos^2 \alpha (1 - \cos^2 \beta) + \sin^2 \alpha \sin^2 \beta \]
Using the identity \( 1 - \cos^2 \beta = \sin^2 \beta \):
\[ A = \cos^2 \alpha \sin^2 \beta + \sin^2 \alpha \sin^2 \beta \]

### Step 6: Factor out \( \sin^2 \beta \)
Now, factor out \( \sin^2 \beta \):
\[ A = \sin^2 \beta (\cos^2 \alpha + \sin^2 \alpha) \]
Using the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \):
\[ A = \sin^2 \beta \]

### Step 7: Determine the range of \( A \)
Since \( \sin^2 \beta \) ranges from 0 to 1 (as \( \beta \) varies), we conclude:
\[ 0 \leq A \leq 1 \]

### Final Answer
Thus, the range of \( A \) is:
\[ [0, 1] \]