Step-by-step Solution:

1. Simplify the function $f(\theta)$

The given function is $f(\theta) = {(\sin \theta + \cos \theta )^2} + {(\sin \theta - \cos \theta )^4}$ for $\theta \in (0,\pi )$.

We use the following trigonometric identities: $(\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = 1 + \sin(2\theta)$ $(\sin \theta - \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta = 1 - \sin(2\theta)$

Substituting these into the expression for $f(\theta)$: $f(\theta) = (1 + \sin(2\theta)) + ((1 - \sin(2\theta))^2)$ Expanding the second term: $f(\theta) = 1 + \sin(2\theta) + (1 - 2\sin(2\theta) + \sin^2(2\theta))$ $f(\theta) = \sin^2(2\theta) - \sin(2\theta) + 2$

2. Find the critical points

To find the local minima, we need to find the critical points by taking the derivative of $f(\theta)$ with respect to $\theta$ and setting it to zero. $f'(\theta) = \frac{d}{{d\theta}} (\sin^2(2\theta) - \sin(2\theta) + 2)$ Using the chain rule: $f'(\theta) = 2\sin(2\theta) \cdot \cos(2\theta) \cdot 2 - \cos(2\theta) \cdot 2$ $f'(\theta) = 4\sin(2\theta)\cos(2\theta) - 2\cos(2\theta)$ $f'(\theta) = 2\cos(2\theta)(2\sin(2\theta) - 1)$

Set $f'(\theta) = 0$ to find the critical points: $2\cos(2\theta)(2\sin(2\theta) - 1) = 0$ This implies either $\cos(2\theta) = 0$ or $2\sin(2 heta) - 1 = 0$.

The domain for $\theta$ is $(0, \pi)$, which means the domain for $2\theta$ is $(0, 2\pi)$.

Case 1: $\cos(2\theta) = 0$ In the interval $(0, 2\pi)$, this occurs when $2\theta = \frac{\pi}{2}$ or $2\theta = \frac{3\pi}{2}$. This gives $\theta = \frac{\pi}{4}$ and $\theta = \frac{3\pi}{4}$.

Case 2: $2\sin(2\theta) - 1 = 0 \implies \sin(2\theta) = \frac{1}{2}$ In the interval $(0, 2\pi)$, this occurs when $2\theta = \frac{\pi}{6}$ or $2\theta = \frac{5\pi}{6}$. This gives $\theta = \frac{\pi}{12}$ and $\theta = \frac{5\pi}{12}$.

So, the critical points in $(0, \pi)$ are $\theta = \frac{\pi}{12}, \frac{\pi}{4}, \frac{5\pi}{12}, \frac{3\pi}{4}$.

3. Classify the critical points

We use the second derivative test. First, we find $f''(\theta)$: $f'(\theta) = 2\sin(4\theta) - 2\cos(2\theta)$ $f''(\theta) = \frac{d}{{d\theta}} (2\sin(4\theta) - 2\cos(2\theta)) = 8\cos(4\theta) + 4\sin(2\theta)$

Now, we evaluate $f''(\theta)$ at each critical point:

• At $\theta = \frac{\pi}{12}$: $2\theta = \frac{\pi}{6}$, $4\theta = \frac{\pi}{3}$. $f''(\frac{\pi}{12}) = 8\cos(\frac{\pi}{3}) + 4\sin(\frac{\pi}{6}) = 8(\frac{1}{2}) + 4(\frac{1}{2}) = 4 + 2 = 6 > 0$. This is a point of local minimum.

• At $\theta = \frac{\pi}{4}$: $2\theta = \frac{\pi}{2}$, $4\theta = \pi$. $f''(\frac{\pi}{4}) = 8\cos(\pi) + 4\sin(\frac{\pi}{2}) = 8(-1) + 4(1) = -4 < 0$. This is a point of local maximum.

• At $\theta = \frac{5\pi}{12}$: $2\theta = \frac{5\pi}{6}$, $4\theta = \frac{5\pi}{3}$. $f''(\frac{5\pi}{12}) = 8\cos(\frac{5\pi}{3}) + 4\sin(\frac{5\pi}{6}) = 8(\frac{1}{2}) + 4(\frac{1}{2}) = 4 + 2 = 6 > 0$. This is a point of local minimum.

• At $\theta = \frac{3\pi}{4}$: $2\theta = \frac{3\pi}{2}$, $4\theta = 3\pi$. $f''(\frac{3\pi}{4}) = 8\cos(3\pi) + 4\sin(\frac{3\pi}{2}) = 8(-1) + 4(-1) = -12 < 0$. This is a point of local maximum.

4. Calculate the required sum

The function $f$ has a local minimum at $\theta$ values $\frac{\pi}{12}$ and $\frac{5\pi}{12}$. The problem states these are given by $\theta \in \{ {\lambda _1}\pi ,....,{\lambda _r}\pi \}$. So, we have $r=2$, with $\lambda_1 \pi = \frac{\pi}{12}$ and $\lambda_2 \pi = \frac{5\pi}{12}$. This gives $\lambda_1 = \frac{1}{12}$ and $\lambda_2 = \frac{5}{12}$. The condition $0 < \lambda_1 < \lambda_2 < 1$ is satisfied as $0 < \frac{1}{12} < \frac{5}{12} < 1$.

The value to be calculated is ${\lambda _1} + ... + {\lambda _r} = \lambda_1 + \lambda_2$. $\lambda_1 + \lambda_2 = \frac{1}{12} + \frac{5}{12} = \frac{6}{12} = \frac{1}{2} = 0.5$

The final value is 0.5.