The maximum value of `sin x + cos x` is equals to ?

Options

A. - sqrt 2
B. sqrt 2
C. 2
D. sqrt 3

Answer

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

### Step 1: Define the function
Let \( y = \sin x + \cos x \).

### Step 2: Differentiate the function
We need to find the derivative of \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos x) = \cos x - \sin x \]

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

### Step 4: Solve for \( x \)
Rearranging gives us:
\[ \cos x = \sin x \]
The angles where sine and cosine are equal occur at:
\[ x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
For the principal value, we take \( x = \frac{\pi}{4} \).

### Step 5: Calculate the maximum value
Now we substitute \( x = \frac{\pi}{4} \) back into the original function:
\[ y = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) \]
Since \( \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \), we have:
\[ y = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} \]

### Step 6: Simplify the result
We can simplify \( \frac{2}{\sqrt{2}} \):
\[ \frac{2}{\sqrt{2}} = \frac{2 \cdot \sqrt{2}}{2} = \sqrt{2} \]

### Conclusion
Thus, the maximum value of \( \sin x + \cos x \) is:
\[ \sqrt{2} \]