Let the function `f : (0, pi) to RR ` be defined by ` f(theta = (sin theta + cos theta)^(2) +(sin theta - cos theta)^(4)` Suppose the function f has a local minimum at `theta` precisely when ` theta in { lambda_(1),pi,....,lambda_(r )pi}` , where `0 lt lambda_(1) lt ....lt lambda_(r ) lt 1 . ` Then the value of `lambda_(1)+....+lambda_(r ) ` is _______

Answer

To solve the problem, we need to analyze the function \( f(\theta) = (\sin \theta + \cos \theta)^2 + (\sin \theta - \cos \theta)^4 \) and find the conditions under which it has local minima.

### Step 1: Simplify the function \( f(\theta) \)

We start by expanding the terms in the function:

1. **First term**:
\[ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 1 + \sin(2\theta) \]

2. **Second term**:
\[ (\sin \theta - \cos \theta)^4 = (\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta)^2 = (1 - 2\sin \theta \cos \theta)^2 = (1 - \sin(2\theta))^2 \]
Expanding this gives:
\[ 1 - 2\sin(2\theta) + \sin^2(2\theta) \]

Combining both terms, we have:
\[ f(\theta) = (1 + \sin(2\theta)) + (1 - 2\sin(2\theta) + \sin^2(2\theta)) = 2 - \sin(2\theta) + \sin^2(2\theta) \]

### Step 2: Find the derivative \( f'(\theta) \)

To find local minima, we need to compute the derivative of \( f(\theta) \):
\[ f'(\theta) = \frac{d}{d\theta}(2 - \sin(2\theta) + \sin^2(2\theta)) \]
Using the chain rule:
\[ f'(\theta) = -2\cos(2\theta) + 2\sin(2\theta)\cos(2\theta) = 2\cos(2\theta)(\sin(2\theta) - 1) \]

### Step 3: Set the derivative to zero

Setting \( f'(\theta) = 0 \):
\[ 2\cos(2\theta)(\sin(2\theta) - 1) = 0 \]
This gives us two cases to consider:
1. \( \cos(2\theta) = 0 \)
2. \( \sin(2\theta) - 1 = 0 \)

### Step 4: Solve for \( \theta \)

1. **From \( \cos(2\theta) = 0 \)**:
\[ 2\theta = \frac{\pi}{2} + n\pi \quad \Rightarrow \quad \theta = \frac{\pi}{4} + \frac{n\pi}{2} \]
In the interval \( (0, \pi) \), valid values are \( \frac{\pi}{4} \) and \( \frac{3\pi}{4} \).

2. **From \( \sin(2\theta) = 1 \)**:
\[ 2\theta = \frac{\pi}{2} + 2k\pi \quad \Rightarrow \quad \theta = \frac{\pi}{4} + k\pi \]
The only valid solution in \( (0, \pi) \) is \( \frac{\pi}{4} \).

### Step 5: Identify \( \lambda_1, \lambda_2, \ldots, \lambda_r \)

From the solutions, we have:
- \( \theta = \frac{\pi}{4} \) corresponds to \( \lambda_1 = \frac{1}{4} \)
- \( \theta = \frac{3\pi}{4} \) corresponds to \( \lambda_2 = \frac{3}{4} \)

### Step 6: Calculate \( \lambda_1 + \lambda_2 \)

Now we sum the values of \( \lambda_1 \) and \( \lambda_2 \):
\[ \lambda_1 + \lambda_2 = \frac{1}{4} + \frac{3}{4} = 1 \]

### Final Answer

Thus, the value of \( \lambda_1 + \lambda_2 \) is:
\[ \boxed{1} \]