Questions: Determine g^(-1)(θ) for g(θ)=1/2 cos(2θ)-3. State the domain and range for both g(θ) and g^(-1)(θ).

Solution Steps

To find the inverse function \( g^{-1}(\theta) \), we need to solve the equation \( y = \frac{1}{2} \cos(2\theta) - 3 \) for \( \theta \) in terms of \( y \). This involves isolating \( \theta \) and expressing it as a function of \( y \). After finding the inverse, we will determine the domain and range of both \( g(\theta) \) and \( g^{-1}(\theta) \).

The function is defined as: \[ g(\theta) = \frac{1}{2} \cos(2\theta) - 3 \] To find the domain and range of \( g(\theta) \), we analyze the cosine function.

The function \( g(\theta) \) is defined for all \( \theta \) in the interval: \[ \theta \in [0, \pi] \] Thus, the domain of \( g(\theta) \) is: \[ \text{Domain of } g(\theta) = [0, \pi] \]

The cosine function oscillates between -1 and 1. Therefore, we calculate the maximum and minimum values of \( g(\theta) \):

• Maximum: \[ g(0) = \frac{1}{2} \cdot 1 - 3 = -2.5 \]
• Minimum: \[ g\left(\frac{\pi}{2}\right) = \frac{1}{2} \cdot (-1) - 3 = -3.5 \] Thus, the range of \( g(\theta) \) is: \[ \text{Range of } g(\theta) = [-3.5, -2.5] \]

To find the inverse function \( g^{-1}(y) \), we solve for \( \theta \): \[ y = \frac{1}{2} \cos(2\theta) - 3 \implies \cos(2\theta) = 2(y + 3) \] Thus, \[ g^{-1}(y) = \frac{1}{2} \arccos(2(y + 3)) \]

The domain of \( g^{-1}(y) \) corresponds to the range of \( g(\theta) \): \[ \text{Domain of } g^{-1}(y) = [-3.5, -2.5] \]

The range of \( g^{-1}(y) \) corresponds to the domain of \( g(\theta) \): \[ \text{Range of } g^{-1}(y) = [0, 1.5692] \]

Final Answer