The value of `(sin x cos y + cos x sin y)^2+(cos x cos y -sin x sin y)^2` is equal to

Answer

To solve the expression \((\sin x \cos y + \cos x \sin y)^2 + (\cos x \cos y - \sin x \sin y)^2\), we can follow these steps:

### Step 1: Recognize the Trigonometric Identities
We can recognize that:
- \(\sin x \cos y + \cos x \sin y\) is the sine addition formula, which simplifies to \(\sin(x + y)\).
- \(\cos x \cos y - \sin x \sin y\) is the cosine addition formula, which simplifies to \(\cos(x + y)\).

### Step 2: Substitute the Identities
Now, we can rewrite the expression using these identities:
\[ (\sin x \cos y + \cos x \sin y)^2 = (\sin(x + y))^2 \]
\[ (\cos x \cos y - \sin x \sin y)^2 = (\cos(x + y))^2 \]
Thus, the original expression becomes:
\[ (\sin(x + y))^2 + (\cos(x + y))^2 \]

### Step 3: Apply the Pythagorean Identity
According to the Pythagorean identity, we know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
for any angle \(\theta\). In our case, \(\theta = x + y\).

### Step 4: Conclude the Result
Therefore, we can conclude that:
\[ (\sin(x + y))^2 + (\cos(x + y))^2 = 1 \]

### Final Answer
The value of the expression \((\sin x \cos y + \cos x \sin y)^2 + (\cos x \cos y - \sin x \sin y)^2\) is equal to **1**.

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