Let `cosA+cosB=x,cos2A+cos2B=y,cos3A+cos3B=z`,then which of the following is true

Options

A. `cos^(2)A+cos^(2)B=1+ y/2`
B. `1/4(2x^(2)-y-2)=cosAcosB`
C. `2x^(2)+z=3x(1+y)`
D. `xyz=0AA A, BepsilonR`

Explanation

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

1. \( x = \cos A + \cos B \)
2. \( y = \cos 2A + \cos 2B \)
3. \( z = \cos 3A + \cos 3B \)

### Step 1: Express \( y \) in terms of \( x \)

Using the double angle formula:
\[ \cos 2A = 2\cos^2 A - 1 \]
\[ \cos 2B = 2\cos^2 B - 1 \]

Thus,
\[ y = \cos 2A + \cos 2B = (2\cos^2 A - 1) + (2\cos^2 B - 1) = 2(\cos^2 A + \cos^2 B) - 2 \]

Now, we can express \( \cos^2 A + \cos^2 B \) in terms of \( x \):
\[ \cos^2 A + \cos^2 B = (\cos A + \cos B)^2 - 2\cos A \cos B = x^2 - 2\cos A \cos B \]

Let \( \cos A \cos B = c \). Then,
\[ y = 2(x^2 - 2c) - 2 = 2x^2 - 4c - 2 \]

### Step 2: Express \( c \) in terms of \( x \) and \( y \)

From the equation for \( y \):
\[ y + 2 = 2x^2 - 4c \implies 4c = 2x^2 - (y + 2) \implies c = \frac{2x^2 - y - 2}{4} \]

### Step 3: Express \( z \) in terms of \( x \) and \( y \)

Using the triple angle formula:
\[ \cos 3A = 4\cos^3 A - 3\cos A \]
\[ \cos 3B = 4\cos^3 B - 3\cos B \]

Thus,
\[ z = \cos 3A + \cos 3B = (4\cos^3 A - 3\cos A) + (4\cos^3 B - 3\cos B) = 4(\cos^3 A + \cos^3 B) - 3(\cos A + \cos B) \]

Using the identity for the sum of cubes:
\[ \cos^3 A + \cos^3 B = (\cos A + \cos B)(\cos^2 A - \cos A \cos B + \cos^2 B) = x\left((\cos^2 A + \cos^2 B) - c\right) \]

Substituting \( \cos^2 A + \cos^2 B = \frac{y + 2}{2} + c \):
\[ z = 4x\left(\frac{y + 2}{2} + c - c\right) - 3x = 2x(y + 2) - 3x = 2xy + 4x - 3x = 2xy + x \]

### Conclusion

We have established the relationships:
1. \( y = 2x^2 - 4c - 2 \)
2. \( c = \frac{2x^2 - y - 2}{4} \)
3. \( z = 2xy + x \)

Now we can analyze which of the given options is true based on these derived equations.