For what value of `theta`, sin `theta + cos theta` is maximum

Answer

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

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

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

### Step 3: Set the derivative to zero
To find the maximum value, we set the derivative equal to zero:
\[ \cos \theta - \sin \theta = 0 \]

### Step 4: Solve for \( \theta \)
From the equation \( \cos \theta = \sin \theta \), we can deduce that:
\[ \tan \theta = 1 \]
This occurs when:
\[ \theta = 45^\circ + n \cdot 180^\circ \quad (n \in \mathbb{Z}) \]
For the principal value, we take \( \theta = 45^\circ \).

### Step 5: Verify that it is a maximum
To confirm that this is a maximum, we can check the second derivative:
\[ \frac{d^2y}{d\theta^2} = -\sin \theta - \cos \theta \]
At \( \theta = 45^\circ \):
\[ \frac{d^2y}{d\theta^2} = -\sin(45^\circ) - \cos(45^\circ) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} < 0 \]
Since the second derivative is negative, \( y \) has a maximum at \( \theta = 45^\circ \).

### Conclusion
Thus, the value of \( \theta \) for which \( \sin \theta + \cos \theta \) is maximum is:
\[ \theta = 45^\circ \]

---