If A ` = 60 ^(@) and B = 30 ^(@) , ` find the value of
`(sin A cos B+ cos A sin B)^(2) + (cos A cos B - sin A sin B )^(2)`

Answer

To solve the expression \( ( \sin A \cos B + \cos A \sin B )^2 + ( \cos A \cos B - \sin A \sin B )^2 \) where \( A = 60^\circ \) and \( B = 30^\circ \), we can follow these steps:

### Step 1: Substitute the values of A and B
We know:
- \( A = 60^\circ \)
- \( B = 30^\circ \)

Now, we can substitute these values into the expression:
\[ ( \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ )^2 + ( \cos 60^\circ \cos 30^\circ - \sin 60^\circ \sin 30^\circ )^2 \]

### Step 2: Calculate the trigonometric values
Using the known values of sine and cosine:
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
- \( \cos 60^\circ = \frac{1}{2} \)
- \( \sin 30^\circ = \frac{1}{2} \)
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)

Substituting these values gives:
\[ ( \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2} )^2 + ( \frac{1}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{1}{2} )^2 \]

### Step 3: Simplify the expression
Now, we simplify each part:
1. For the first part:
\[ \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3}{4} \]
\[ \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \]
Thus,
\[ \frac{3}{4} + \frac{1}{4} = 1 \]
So,
\[ (1)^2 = 1 \]

2. For the second part:
\[ \frac{1}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = 0 \]
Thus,
\[ (0)^2 = 0 \]

### Step 4: Combine the results
Now, we combine the results from both parts:
\[ 1 + 0 = 1 \]

### Final Answer
The value of the expression is:
\[ \boxed{1} \]