If x = 3 sin `theta + 4 cos theta and y = 3 cos theta - 4 sin theta` then prove that `x^(2) + y^(2) = 25`.

Answer

To prove that \( x^2 + y^2 = 25 \) given \( x = 3 \sin \theta + 4 \cos \theta \) and \( y = 3 \cos \theta - 4 \sin \theta \), we will follow these steps:

### Step 1: Calculate \( x^2 \)
We start with the expression for \( x \):
\[ x = 3 \sin \theta + 4 \cos \theta \]
Now, we square \( x \):
\[ x^2 = (3 \sin \theta + 4 \cos \theta)^2 \]
Using the formula \( (a + b)^2 = a^2 + b^2 + 2ab \):
\[ x^2 = (3 \sin \theta)^2 + (4 \cos \theta)^2 + 2(3 \sin \theta)(4 \cos \theta) \]
Calculating each term:
\[ x^2 = 9 \sin^2 \theta + 16 \cos^2 \theta + 24 \sin \theta \cos \theta \]

### Step 2: Calculate \( y^2 \)
Next, we take the expression for \( y \):
\[ y = 3 \cos \theta - 4 \sin \theta \]
Now, we square \( y \):
\[ y^2 = (3 \cos \theta - 4 \sin \theta)^2 \]
Using the formula \( (a - b)^2 = a^2 + b^2 - 2ab \):
\[ y^2 = (3 \cos \theta)^2 + (-4 \sin \theta)^2 - 2(3 \cos \theta)(4 \sin \theta) \]
Calculating each term:
\[ y^2 = 9 \cos^2 \theta + 16 \sin^2 \theta - 24 \cos \theta \sin \theta \]

### Step 3: Add \( x^2 \) and \( y^2 \)
Now we add \( x^2 \) and \( y^2 \):
\[ x^2 + y^2 = (9 \sin^2 \theta + 16 \cos^2 \theta + 24 \sin \theta \cos \theta) + (9 \cos^2 \theta + 16 \sin^2 \theta - 24 \sin \theta \cos \theta) \]
Combining like terms:
\[ x^2 + y^2 = (9 \sin^2 \theta + 16 \sin^2 \theta) + (16 \cos^2 \theta + 9 \cos^2 \theta) + (24 \sin \theta \cos \theta - 24 \sin \theta \cos \theta) \]
The \( 24 \sin \theta \cos \theta \) terms cancel out:
\[ x^2 + y^2 = 25 \sin^2 \theta + 25 \cos^2 \theta \]

### Step 4: Use the Pythagorean Identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ x^2 + y^2 = 25(\sin^2 \theta + \cos^2 \theta) = 25 \cdot 1 = 25 \]

### Conclusion
Thus, we have proved that:
\[ x^2 + y^2 = 25 \]