The maximum value of `cos theta + sin theta` , if

Options

A. `theta = 45^@`
B. `theta = 60^@`
C. `theta = 90^@`
D. `theta = 30^@`

Answer

To find the maximum value of \( \cos \theta + \sin \theta \), we can follow these steps:

### Step 1: Set up the function
Let \( f(\theta) = \cos \theta + \sin \theta \).

### Step 2: Find the derivative
To find the maximum value, we need to take the derivative of \( f(\theta) \):
\[ f'(\theta) = \frac{d}{d\theta}(\cos \theta + \sin \theta) = -\sin \theta + \cos \theta \]

### Step 3: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ -\sin \theta + \cos \theta = 0 \]
This simplifies to:
\[ \cos \theta = \sin \theta \]

### Step 4: Solve for \( \theta \)
Dividing both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ 1 = \tan \theta \]
This implies:
\[ \tan \theta = 1 \]
The angle \( \theta \) that satisfies this is:
\[ \theta = 45^\circ \quad \text{(or } \frac{\pi}{4} \text{ radians)} \]

### Step 5: Find the maximum value
To find the maximum value of \( f(\theta) \), we can substitute \( \theta = 45^\circ \) back into the original function:
\[ f(45^\circ) = \cos(45^\circ) + \sin(45^\circ) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Conclusion
Thus, the maximum value of \( \cos \theta + \sin \theta \) is \( \sqrt{2} \) when \( \theta = 45^\circ \).

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