If `a cos theta - b sin theta = lambda`, then `AA` real `theta`

Options

A. `lambda gt sqrt( a^(2) + b^(2))`
B. `lambda lt- sqrt( a^(2) + b^(2))`
C. `- sqrt( a^(2) +b^(2)) le lambdale sqrt( a^(2) +b^(2))`
D. none

Answer

To solve the problem \( a \cos \theta - b \sin \theta = \lambda \) and find the maximum and minimum values of \( \lambda \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \lambda = a \cos \theta - b \sin \theta \]

### Step 2: Factor out the square root of the sum of squares
To analyze the expression, we can factor out \( \sqrt{a^2 + b^2} \):
\[ \lambda = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \cos \theta - \frac{b}{\sqrt{a^2 + b^2}} \sin \theta \right) \]

### Step 3: Introduce trigonometric identities
Let:
\[ \sin x = \frac{a}{\sqrt{a^2 + b^2}} \quad \text{and} \quad \cos x = \frac{b}{\sqrt{a^2 + b^2}} \]
This means:
\[ \lambda = \sqrt{a^2 + b^2} \left( \sin x \cos \theta - \cos x \sin \theta \right) \]

### Step 4: Use the sine subtraction formula
Using the sine subtraction formula \( \sin(A - B) = \sin A \cos B - \cos A \sin B \):
\[ \lambda = \sqrt{a^2 + b^2} \sin(x - \theta) \]

### Step 5: Determine the range of \( \sin \)
The sine function has a range of:
\[ -1 \leq \sin(x - \theta) \leq 1 \]
Thus, we can write:
\[ -\sqrt{a^2 + b^2} \leq \lambda \leq \sqrt{a^2 + b^2} \]

### Conclusion
The minimum value of \( \lambda \) is \( -\sqrt{a^2 + b^2} \) and the maximum value is \( \sqrt{a^2 + b^2} \).