If an angle B is complement of an angle A, what are the greatest and last values of cos A cos B respectively ?

Options

A. `0,-(1)/(2)`
B. `(1)/(2),-1`
C. `1 , 0`
D. `(1)/(2),(1)/(2)`

Explanation

To solve the problem of finding the greatest and least values of \( \cos A \cos B \) where angle \( B \) is the complement of angle \( A \), we can follow these steps:

### Step 1: Understand the relationship between angles A and B
Since \( B \) is the complement of \( A \), we have:
\[ A + B = 90^\circ \quad \text{or} \quad B = 90^\circ - A \]

### Step 2: Rewrite the expression
We need to express \( \cos A \cos B \) in terms of \( A \):
\[ \cos B = \cos(90^\circ - A) = \sin A \]
Thus, we can rewrite the expression:
\[ \cos A \cos B = \cos A \sin A \]

### Step 3: Use a trigonometric identity
We can use the double angle identity for sine:
\[ \sin 2A = 2 \sin A \cos A \]
This means:
\[ \cos A \sin A = \frac{1}{2} \sin 2A \]

### Step 4: Determine the range of \( \sin 2A \)
The sine function oscillates between -1 and 1. Therefore, the range of \( \sin 2A \) is:
\[ -1 \leq \sin 2A \leq 1 \]

### Step 5: Find the range of \( \frac{1}{2} \sin 2A \)
Multiplying the entire range of \( \sin 2A \) by \( \frac{1}{2} \):
\[ -\frac{1}{2} \leq \frac{1}{2} \sin 2A \leq \frac{1}{2} \]

### Step 6: Identify the greatest and least values
From the above inequality, we can conclude:
- The least value of \( \cos A \cos B \) is \( -\frac{1}{2} \)
- The greatest value of \( \cos A \cos B \) is \( \frac{1}{2} \)

### Final Answer
The greatest and least values of \( \cos A \cos B \) are \( \frac{1}{2} \) and \( -\frac{1}{2} \) respectively.

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