Suppose that 'a' is a non-zero real number for which `sin x+sin y=a and cos x + cos y= 2a`. The value of `cos(x-y) is`

Options

A. `(3a^(2)-2)/(2)`
B. `(7a^(2)-2)/(2)`
C. `(9a^(2)-2)/ (2)`
D. `(5a^(2)-2)/(2)`

Answer

To solve the problem, we need to find the value of \( \cos(x - y) \) given the equations:

1. \( \sin x + \sin y = a \)
2. \( \cos x + \cos y = 2a \)

### Step 1: Square both equations
We start by squaring both equations:

\[ (\sin x + \sin y)^2 = a^2 \]
\[ (\cos x + \cos y)^2 = (2a)^2 = 4a^2 \]

### Step 2: Expand the squared equations
Now, we expand both squared equations:

\[ \sin^2 x + 2\sin x \sin y + \sin^2 y = a^2 \]
\[ \cos^2 x + 2\cos x \cos y + \cos^2 y = 4a^2 \]

### Step 3: Combine the equations
Next, we add the two expanded equations:

\[ (\sin^2 x + \cos^2 x) + (\sin^2 y + \cos^2 y) + 2(\sin x \sin y + \cos x \cos y) = a^2 + 4a^2 \]

Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) and \( \sin^2 y + \cos^2 y = 1 \):

\[ 1 + 1 + 2(\sin x \sin y + \cos x \cos y) = 5a^2 \]

This simplifies to:

\[ 2 + 2(\sin x \sin y + \cos x \cos y) = 5a^2 \]

### Step 4: Isolate the cosine term
Now, we isolate \( \sin x \sin y + \cos x \cos y \):

\[ 2(\sin x \sin y + \cos x \cos y) = 5a^2 - 2 \]

Dividing both sides by 2 gives:

\[ \sin x \sin y + \cos x \cos y = \frac{5a^2 - 2}{2} \]

### Step 5: Use the cosine of difference identity
Recall the cosine of the difference identity:

\[ \cos(x - y) = \cos x \cos y + \sin x \sin y \]

Thus, we can substitute:

\[ \cos(x - y) = \frac{5a^2 - 2}{2} \]

### Final Answer
The value of \( \cos(x - y) \) is:

\[ \cos(x - y) = \frac{5a^2 - 2}{2} \]

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