If `sin x+siny=a, cosx+cosy=b`, then what is the value of `cos(x-y)`?

Options

A. `a^(2)-1`
B. `b^(2)-1`
C. `(1)/(2)(a^(2)+b^(2)-2)`
D. `(1)/(2)(a^(2)+b^(2))`

Explanation

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

### Step 1: Square both equations
Start by squaring both equations:
\[ (\sin x + \sin y)^2 = a^2 \quad \text{(1)} \]
\[ (\cos x + \cos y)^2 = b^2 \quad \text{(2)} \]

### Step 2: Expand the squared equations
Expand both squared equations:
\[ \sin^2 x + 2 \sin x \sin y + \sin^2 y = a^2 \quad \text{(from 1)} \]
\[ \cos^2 x + 2 \cos x \cos y + \cos^2 y = b^2 \quad \text{(from 2)} \]

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

### Step 4: Use the Pythagorean identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can simplify:
\[ 1 + 1 + 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2 \]
This simplifies to:
\[ 2 + 2(\sin x \sin y + \cos x \cos y) = a^2 + b^2 \]

### Step 5: Isolate the cosine of the difference
Now, recall the cosine of the difference identity:
\[ \cos(x-y) = \cos x \cos y + \sin x \sin y \]
Thus, we can rewrite our equation:
\[ 2 + 2\cos(x-y) = a^2 + b^2 \]
Dividing through by 2 gives:
\[ 1 + \cos(x-y) = \frac{a^2 + b^2}{2} \]

### Step 6: Solve for \( \cos(x-y) \)
Now, isolate \( \cos(x-y) \):
\[ \cos(x-y) = \frac{a^2 + b^2}{2} - 1 \]

### Final Result
Thus, the value of \( \cos(x-y) \) is:
\[ \cos(x-y) = \frac{a^2 + b^2}{2} - 1 \]

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