Let $$f(\theta) = 3\left(\sin^4\left(\frac{3\pi}{2} - \theta\right) + \sin^4(3\pi + \theta)\right) - 2\left(1 - \sin^2 2\theta\right)$$ and $$S = \left\{\theta \in [0, \pi] : f'(\theta) = -\frac{\sqrt{3}}{2}\right\}$$. If $$4\beta = \sum_{\theta \in S} \theta$$ then $$f(\beta)$$ is equal to

Step 1: Simplifying the Trigonometric Function $$f(\theta)$$

The original function is given by:

$$f(\theta) = 3 \left( \sin^4\left(\frac{3\pi}{2} - \theta\right) + \sin^4(3\pi + \theta) \right) - 2(1 - \sin^2 2\theta)$$

First, simplify the terms inside the powers using allied angle properties:

• $$\sin\left(\frac{3\pi}{2} - \theta\right) = -\cos\theta$$

  (Third quadrant, sine is negative)

• $$\sin(3\pi + \theta) = \sin(\pi + \theta) = -\sin\theta$$

  (Third quadrant, sine is negative)

• $$1 - \sin^2 2\theta = \cos^2 2\theta$$

  (Fundamental identity)

Substitute these back into the function:

$$f(\theta) = 3 \left( (-\cos\theta)^4 + (-\sin\theta)^4 \right) - 2\cos^2 2\theta$$
$$f(\theta) = 3 (\cos^4\theta + \sin^4\theta) - 2\cos^2 2\theta$$

Now, apply the algebraic identity

$$a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2$$ :

$$\cos^4\theta + \sin^4\theta = (\cos^2\theta + \sin^2\theta)^2 - 2\sin^2\theta\cos^2\theta$$
$$= 1 - 2\sin^2\theta\cos^2\theta$$

Using the double angle formula

$$\sin 2\theta = 2\sin\theta\cos\theta$$, we can write

$$2\sin^2\theta\cos^2\theta = \frac{1}{2}\sin^2 2\theta$$.

So, $$\cos^4\theta + \sin^4\theta = 1 - \frac{1}{2}\sin^2 2\theta$$

Substitute this back into $$f(\theta)$$ :

$$f(\theta) = 3 \left( 1 - \frac{1}{2}\sin^2 2\theta \right) - 2(1 - \sin^2 2\theta)$$
$$f(\theta) = 3 - \frac{3}{2}\sin^2 2\theta - 2 + 2\sin^2 2\theta$$
$$f(\theta) = 1 + \frac{1}{2}\sin^2 2\theta$$

Step 2: Finding the Derivative $$f'(\theta)$$

Differentiate the simplified

$$f(\theta)$$with respect to$$\theta$$

using the chain rule:

$$f'(\theta) = \frac{d}{d\theta} \left( 1 + \frac{1}{2}\sin^2 2\theta \right)$$
$$f'(\theta) = \frac{1}{2} \cdot 2\sin 2\theta \cdot \cos 2\theta \cdot 2$$
$$f'(\theta) = 2\sin 2\theta\cos 2\theta$$

Using the double angle identity again ($$\sin 2A = 2\sin A\cos A$$), we get:

$$f'(\theta) = \sin 4\theta$$

Step 3: Solving the Trigonometric Equation for Set $S$

We are given that

$$S = \left\{ \theta \in [0, \pi] : f'(\theta) = -\frac{\sqrt{3}}{2} \right\}$$.

$$\sin 4\theta = -\frac{\sqrt{3}}{2}$$

Since$$\theta \in [0, \pi]$$, the interval for $$4\theta$$ is$$[0, 4\pi]$$.

We need to find all angles in

$$[0, 4\pi]$$where the sine value is$$-\frac{\sqrt{3}}{2}$$.
These will be in the third and fourth quadrants of each cycle.

For the first cycle

$$[0, 2\pi]$$ :

$$4\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$$
$$4\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$$

For the second cycle

$$[2\pi, 4\pi]$$ :

$$4\theta = 3\pi + \frac{\pi}{3} = \frac{10\pi}{3}$$
$$4\theta = 4\pi - \frac{\pi}{3} = \frac{11\pi}{3}$$

The valid values for $$\theta$$ are:

$$S = \left\{ \frac{\pi}{3}, \frac{5\pi}{12}, \frac{5\pi}{6}, \frac{11\pi}{12} \right\}$$

Step 4: Finding $$4\beta$$ and Evaluating $$f(\beta)$$

We are given

$$4\beta = \sum_{\theta \in S} \theta$$.

Let's sum the elements of set $$S$$ :

$$4\beta = \frac{\pi}{3} + \frac{5\pi}{12} + \frac{5\pi}{6} + \frac{11\pi}{12}$$

Convert all fractions to a common denominator of 12:

$$4\beta = \frac{4\pi}{12} + \frac{5\pi}{12} + \frac{10\pi}{12} + \frac{11\pi}{12}$$
$$4\beta = \frac{30\pi}{12}$$
$$4\beta = \frac{5\pi}{2} \implies \beta = \frac{5\pi}{8}$$

Finally, evaluate $$f(\beta)$$using our simplified function$$f(\theta) = 1 + \frac{1}{2}\sin^2 2\theta$$ :

$$f(\beta) = 1 + \frac{1}{2}\sin^2\left(2 \cdot \frac{5\pi}{8}\right)$$
$$f(\beta) = 1 + \frac{1}{2}\sin^2\left(\frac{5\pi}{4}\right)$$

We know that

$$\sin\left(\frac{5\pi}{4}\right) = \sin\left(\pi + \frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}$$.

$$f(\beta) = 1 + \frac{1}{2}\left(-\frac{1}{\sqrt{2}}\right)^2$$
$$f(\beta) = 1 + \frac{1}{2} \left( \frac{1}{2} \right)$$
$$f(\beta) = 1 + \frac{1}{4}$$
$$f(\beta) = \frac{5}{4}$$

Hence, the correct answer is Option B.