If `x = a cos theta - b sin theta and y = b cos theta + a sin theta`, then the value of `x^(2) + y^(2)` is

Options

A. `a^2`
B. `b^2`
C. `(a^2)/(b^2)`
D. `a^2 + b^2`

Answer

To find the value of \( x^2 + y^2 \) given \( x = a \cos \theta - b \sin \theta \) and \( y = b \cos \theta + a \sin \theta \), we can follow these steps:

### Step 1: Calculate \( x^2 \)
We start with the expression for \( x \):
\[ x = a \cos \theta - b \sin \theta \]
Now, squaring \( x \):
\[ x^2 = (a \cos \theta - b \sin \theta)^2 \]
Using the formula \( (p - q)^2 = p^2 - 2pq + q^2 \):
\[ x^2 = a^2 \cos^2 \theta - 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta \]

### Step 2: Calculate \( y^2 \)
Next, we calculate \( y^2 \):
\[ y = b \cos \theta + a \sin \theta \]
Squaring \( y \):
\[ y^2 = (b \cos \theta + a \sin \theta)^2 \]
Using the formula \( (p + q)^2 = p^2 + 2pq + q^2 \):
\[ y^2 = b^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + a^2 \sin^2 \theta \]

### Step 3: Add \( x^2 \) and \( y^2 \)
Now, we add \( x^2 \) and \( y^2 \):
\[ x^2 + y^2 = (a^2 \cos^2 \theta - 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta) + (b^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + a^2 \sin^2 \theta) \]

### Step 4: Combine like terms
Combining the terms:
\[ x^2 + y^2 = a^2 \cos^2 \theta + b^2 \cos^2 \theta + a^2 \sin^2 \theta + b^2 \sin^2 \theta \]
This can be rewritten as:
\[ x^2 + y^2 = (a^2 + b^2)(\cos^2 \theta + \sin^2 \theta) \]

### Step 5: Use the Pythagorean identity
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ x^2 + y^2 = (a^2 + b^2) \cdot 1 \]
Thus, we conclude:
\[ x^2 + y^2 = a^2 + b^2 \]

### Final Answer
The value of \( x^2 + y^2 \) is:
\[ \boxed{a^2 + b^2} \]