If sin x + sin y = a, cos x + cos y = 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)`

Answer

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

### Step 1: Square the given equations
We start by squaring both equations.

1. From \( \sin x + \sin y = a \):
\[ (\sin x + \sin y)^2 = a^2 \]
Expanding this gives:
\[ \sin^2 x + \sin^2 y + 2 \sin x \sin y = a^2 \]
This is our first equation.

2. From \( \cos x + \cos y = b \):
\[ (\cos x + \cos y)^2 = b^2 \]
Expanding this gives:
\[ \cos^2 x + \cos^2 y + 2 \cos x \cos y = b^2 \]
This is our second equation.

### Step 2: Add the two equations
Now, we add both equations:
\[ (\sin^2 x + \sin^2 y + 2 \sin x \sin y) + (\cos^2 x + \cos^2 y + 2 \cos x \cos y) = a^2 + b^2 \]
This simplifies to:
\[ (\sin^2 x + \cos^2 x) + (\sin^2 y + \cos^2 y) + 2 (\sin x \sin y + \cos x \cos y) = a^2 + b^2 \]

### Step 3: Use the Pythagorean identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 1 + 1 + 2 (\sin x \sin y + \cos x \cos y) = a^2 + b^2 \]
This simplifies to:
\[ 2 + 2 (\sin x \sin y + \cos x \cos y) = a^2 + b^2 \]

### Step 4: Isolate \( \sin x \sin y + \cos x \cos y \)
Now, we can 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:
\[ \sin x \sin y + \cos x \cos y = \frac{a^2 + b^2 - 2}{2} \]

### Step 5: Use the cosine difference identity
Recall that:
\[ \cos(x - y) = \cos x \cos y + \sin x \sin y \]
Thus, we can substitute our result:
\[ \cos(x - y) = \frac{a^2 + b^2 - 2}{2} \]

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

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