Prove that `(sin alpha cos beta + cos alpha sin beta) ^(2) + (cos alpha coa beta - sin alpha sin beta) ^(2) =1.`

Answer

To prove that

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

we will start by simplifying the left-hand side (LHS) using trigonometric identities.

### Step 1: Identify the terms

We can identify the two terms in the LHS:

1. \( A = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)
2. \( B = \cos \alpha \cos \beta - \sin \alpha \sin \beta \)

### Step 2: Recognize the sine and cosine of sum formulas

We know from trigonometric identities that:

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

Thus, we can rewrite \( A \) and \( B \) as:

\[ A = \sin(\alpha + \beta) \]
\[ B = \cos(\alpha + \beta) \]

### Step 3: Substitute back into the LHS

Now substituting \( A \) and \( B \) into the LHS gives us:

\[ LHS = (\sin(\alpha + \beta))^2 + (\cos(\alpha + \beta))^2 \]

### Step 4: Apply the Pythagorean identity

Using the Pythagorean identity, we know that:

\[ \sin^2 x + \cos^2 x = 1 \]

Thus, we have:

\[ LHS = 1 \]

### Conclusion

We have shown that:

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

Therefore, the statement is proved.