If `y + "cos" theta = "sin" theta` has a real solution, then

Options

A. `-sqrt(2) le y le sqrt(2)`
B. `y gt sqrt(2)`
C. `y le - sqrt(2)`
D. none of these

Answer

To solve the equation \( y + \cos \theta = \sin \theta \) and determine the conditions under which it has a real solution, we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given equation:
\[ y + \cos \theta = \sin \theta \]
This can be rewritten as:
\[ y = \sin \theta - \cos \theta \]

### Step 2: Expressing in a Different Form
Next, we can factor out \(\sqrt{2}\) from the right-hand side. To do this, we can multiply and divide by \(\sqrt{2}\):
\[ y = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \theta - \frac{1}{\sqrt{2}} \cos \theta \right) \]

### Step 3: Recognizing Trigonometric Identities
We know that:
\[ \frac{1}{\sqrt{2}} = \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) \]
Thus, we can rewrite the equation as:
\[ y = \sqrt{2} \left( \sin \theta \cos\left(\frac{\pi}{4}\right) - \cos \theta \sin\left(\frac{\pi}{4}\right) \right) \]
This is equivalent to:
\[ y = \sqrt{2} \sin\left(\theta - \frac{\pi}{4}\right) \]

### Step 4: Finding the Range of \(y\)
The sine function has a range of \([-1, 1]\). Therefore:
\[ \sqrt{2} \sin\left(\theta - \frac{\pi}{4}\right) \text{ will range from } -\sqrt{2} \text{ to } \sqrt{2}. \]
This implies:
\[ -\sqrt{2} \leq y \leq \sqrt{2} \]

### Step 5: Conclusion
For the equation \( y + \cos \theta = \sin \theta \) to have a real solution, \( y \) must satisfy:
\[ -\sqrt{2} \leq y \leq \sqrt{2} \]