If `x +y= 3-cos4theta` and `x-y=4sin2theta` then

Options

A. `x^(4)+y^(4)=9`
B. `sqrt(x)+sqrt(y)=16`
C. `x^(3)+y^(3)=2(x^(2)+y^(2))`
D. `sqrt(x)+sqrt(y)=2`

Answer

To solve the equations given in the problem, we will follow these steps:

### Step 1: Write down the equations
We have two equations:
1. \( x + y = 3 - \cos(4\theta) \) (Equation 1)
2. \( x - y = 4\sin(2\theta) \) (Equation 2)

### Step 2: Solve for \( x \) and \( y \)
To find \( x \) and \( y \), we can add and subtract the two equations.

**Adding Equation 1 and Equation 2:**
\[ (x + y) + (x - y) = (3 - \cos(4\theta)) + (4\sin(2\theta)) \]
This simplifies to:
\[ 2x = 3 - \cos(4\theta) + 4\sin(2\theta) \]
Thus, we can express \( x \) as:
\[ x = \frac{3 - \cos(4\theta) + 4\sin(2\theta)}{2} \quad \text{(Equation 3)} \]

**Subtracting Equation 2 from Equation 1:**
\[ (x + y) - (x - y) = (3 - \cos(4\theta)) - (4\sin(2\theta)) \]
This simplifies to:
\[ 2y = 3 - \cos(4\theta) - 4\sin(2\theta) \]
Thus, we can express \( y \) as:
\[ y = \frac{3 - \cos(4\theta) - 4\sin(2\theta)}{2} \quad \text{(Equation 4)} \]

### Step 3: Substitute \( x \) and \( y \) into a new equation
Now, we can use the values of \( x \) and \( y \) to form a new equation. We can use the identity \( x^2 + y^2 = (x+y)^2 - 2xy \).

From Equation 1:
\[ x + y = 3 - \cos(4\theta) \]
From Equation 2:
\[ x - y = 4\sin(2\theta) \]

### Step 4: Find \( x^2 + y^2 \)
Using the identity:
\[ x^2 + y^2 = (x+y)^2 - 2xy \]
We need to find \( xy \). We can find \( xy \) using:
\[ xy = \frac{(x+y)^2 - (x-y)^2}{4} \]

Calculating \( (x+y)^2 \) and \( (x-y)^2 \):
\[ (x+y)^2 = (3 - \cos(4\theta))^2 \]
\[ (x-y)^2 = (4\sin(2\theta))^2 \]

Substituting these into the equation for \( xy \):
\[ xy = \frac{(3 - \cos(4\theta))^2 - (4\sin(2\theta))^2}{4} \]

### Step 5: Final equation
Now, we can express the final equation in terms of \( x \) and \( y \):
\[ x^2 + y^2 - 8x - 8y + 16 = 0 \]

This gives us the final result.