If `(sinx + sin y ) = a` and `(cos x + cos y ) = b ` what is the value of sin x sin y + cos x cos y ?

Options

A. `a + b - ab `
B. `a + b+ ab`
C. `a^(2) + b^(2) - 2 `
D. ` (a^(2) + b^(2) - 2)/(2)`

Answer

To find the value of \( \sin x \sin y + \cos x \cos y \) given that \( \sin x + \sin y = a \) and \( \cos x + \cos y = b \), we can follow these steps:

### Step 1: Square the equations
We start with the two equations:
1. \( \sin x + \sin y = a \)
2. \( \cos x + \cos y = b \)

We square both equations.

\[ (\sin x + \sin y)^2 = a^2 \implies \sin^2 x + \sin^2 y + 2 \sin x \sin y = a^2 \tag{1} \]

\[ (\cos x + \cos y)^2 = b^2 \implies \cos^2 x + \cos^2 y + 2 \cos x \cos y = b^2 \tag{2} \]

### Step 2: Use the Pythagorean identity
From the Pythagorean identity, we know that:
\[ \sin^2 x + \cos^2 x = 1 \quad \text{and} \quad \sin^2 y + \cos^2 y = 1 \]

Adding these identities gives:
\[ (\sin^2 x + \cos^2 x) + (\sin^2 y + \cos^2 y) = 1 + 1 = 2 \]

### Step 3: Combine the squared equations
Now, we add equations (1) and (2):
\[ (\sin^2 x + \sin^2 y + 2 \sin x \sin y) + (\cos^2 x + \cos^2 y + 2 \cos x \cos y) = a^2 + b^2 \]

This simplifies to:
\[ (\sin^2 x + \cos^2 x) + (\sin^2 y + \cos^2 y) + 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2 \]

Substituting \( \sin^2 x + \cos^2 x + \sin^2 y + \cos^2 y = 2 \):
\[ 2 + 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2 \]

### Step 4: Solve for \( \sin x \sin y + \cos x \cos y \)
Now, we isolate \( \sin x \sin y + \cos x \cos y \):
\[ 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2 - 2 \]

Dividing both sides by 2 gives:
\[ \sin x \sin y + \cos x \cos y = \frac{a^2 + b^2 - 2}{2} \]

### Final Answer
Thus, the value of \( \sin x \sin y + \cos x \cos y \) is:
\[ \sin x \sin y + \cos x \cos y = \frac{a^2 + b^2 - 2}{2} \]

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