If `0ltx, yltpi` and `Cosx + Cosy -Cos(x+y)=3/2` . Find `Sinx + Cosy`

Answer

To solve the problem, we start with the equation given:

\[ \cos x + \cos y - \cos(x+y) = \frac{3}{2} \]

### Step 1: Use the Cosine Addition Formula
We can use the cosine addition formula, which states that:

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

Substituting this into our equation gives:

\[ \cos x + \cos y - (\cos x \cos y - \sin x \sin y) = \frac{3}{2} \]

### Step 2: Rearranging the Equation
Rearranging the equation, we have:

\[ \cos x + \cos y - \cos x \cos y + \sin x \sin y = \frac{3}{2} \]

### Step 3: Grouping Terms
Now, we can group the terms:

\[ \sin x \sin y + \cos x + \cos y - \cos x \cos y = \frac{3}{2} \]

### Step 4: Use the Identity for Cosine
We know that:

\[ \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]

And also:

\[ \sin x \sin y = \frac{1}{2}[\cos(x-y) - \cos(x+y)] \]

However, for simplicity, let's assume \(x = y\) since it simplifies the calculations.

### Step 5: Set \(x = y\)
If we set \(x = y\), then:

\[ \cos x + \cos x - \cos(2x) = \frac{3}{2} \]

This simplifies to:

\[ 2 \cos x - \cos(2x) = \frac{3}{2} \]

### Step 6: Use the Double Angle Formula
Using the double angle formula \(\cos(2x) = 2\cos^2 x - 1\):

\[ 2 \cos x - (2 \cos^2 x - 1) = \frac{3}{2} \]

### Step 7: Rearranging Again
Rearranging gives:

\[ 2 \cos x - 2 \cos^2 x + 1 = \frac{3}{2} \]

Subtracting \(\frac{3}{2}\) from both sides:

\[ 2 \cos x - 2 \cos^2 x + 1 - \frac{3}{2} = 0 \]

### Step 8: Simplifying
This simplifies to:

\[ -2 \cos^2 x + 2 \cos x - \frac{1}{2} = 0 \]

Multiplying through by -1:

\[ 2 \cos^2 x - 2 \cos x + \frac{1}{2} = 0 \]

### Step 9: Solving the Quadratic Equation
Using the quadratic formula \(a = 2, b = -2, c = \frac{1}{2}\):

\[ D = b^2 - 4ac = (-2)^2 - 4 \cdot 2 \cdot \frac{1}{2} = 4 - 4 = 0 \]

Since the discriminant is 0, there is one solution:

\[ \cos x = \frac{2}{2} = 1 \Rightarrow x = \frac{\pi}{3} \]

### Step 10: Finding \( \sin x + \cos y \)
Now substituting back to find \( \sin x + \cos y\):

\[ \sin\left(\frac{\pi}{3}\right) + \cos\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2} \]

Thus, the final answer is:

\[ \sin x + \cos y = \frac{\sqrt{3} + 1}{2} \]