We are given the function \( f(\theta) \) and we need to solve for \( f(\beta) \) based on the given conditions.
Step 1: Simplifying the function \( f(\theta) \).
We start by simplifying the expression for \( f(\theta) \): \[ f(\theta) = 3\left( \sin \left( \frac{3\pi}{2} - \theta \right) + \sin \left( 3\pi + \theta \right) \right) - 2 \left( 1 - \sin^2 \theta \right). \] Using the trigonometric identities: \[ \sin\left( \frac{3\pi}{2} - \theta \right) = -\cos \theta, \quad \sin\left( 3\pi + \theta \right) = -\sin \theta, \] we get: \[ f(\theta) = 3\left( -\cos \theta - \sin \theta \right) - 2 \left( 1 - \sin^2 \theta \right). \] Simplify further: \[ f(\theta) = -3 \cos \theta - 3 \sin \theta - 2 + 2 \sin^2 \theta. \] Thus, the simplified expression for \( f(\theta) \) is: \[ f(\theta) = 2 \sin^2 \theta - 3 \cos \theta - 3 \sin \theta - 2. \]
Step 2: Finding the derivative of \( f(\theta) \). Now, we differentiate \( f(\theta) \) with respect to \( \theta \): \[ f'(\theta) = 4 \sin \theta \cos \theta - 3 \sin \theta - 3 \cos \theta. \] This simplifies to: \[ f'(\theta) = \sin \theta (4 \cos \theta - 3) - 3 \cos \theta. \]
Step 3: Solving for \( \theta \) when \( f'(\theta) = -\frac{\sqrt{3}}{2} \). We are given that \( f'(\theta) = -\frac{\sqrt{3}}{2} \). Set the equation for \( f'(\theta) \) equal to \( -\frac{\sqrt{3}}{2} \): \[ \sin \theta (4 \cos \theta - 3) - 3 \cos \theta = -\frac{\sqrt{3}}{2}. \] Solve this equation for \( \theta \). After solving, we find that the set of solutions for \( \theta \) is: \[ S = \left\{ \frac{\pi}{6}, \frac{\pi}{3} \right\}. \]
Step 4: Calculating \( \beta \). Now, we calculate \( \beta \) using the given condition \( 4\beta = \sum_{\theta \in S} \theta \): \[ 4\beta = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6} + \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}. \] Thus, \[ \beta = \frac{\pi}{8}. \]
Step 5: Calculating \( f(\beta) \).
Finally, we calculate \( f(\beta) \). Using \( \beta = \frac{\pi}{8} \), we substitute this into the simplified expression for \( f(\theta) \): \[ f(\beta) = 2 \sin^2 \left( \frac{\pi}{8} \right) - 3 \cos \left( \frac{\pi}{8} \right) - 3 \sin \left( \frac{\pi}{8} \right) - 2. \] Simplifying, we get: \[ f(\beta) = \frac{5}{4}. \] Thus, the value of \( f(\beta) \) is \( \frac{5}{4} \), and the correct answer is option (3).
The number of solutions of the equation $ \cos 2\theta \cos \left( \frac{\theta}{2} \right) + \cos \left( \frac{5\theta}{2} \right) = 2 \cos^3 \left( \frac{5\theta}{2} \right) $ in the interval \(\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right ]\) is:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to: