1. **Problem Statement:** Determine the values of scalars $A$ and $\omega$ for the trigonometric function $$y = A \sin(\omega x + 30^\circ)$$ given the graph of the wave for $$-90^\circ \leq x \leq 90^\circ$$. 2. **Understanding the function:** The general form is $$y = A \sin(\omega x + \phi)$$ where: - $A$ is the amplitude (maximum absolute value of $y$). - $\omega$ is the angular frequency, related to the period $T$ by $$T = \frac{360^\circ}{\omega}$$ since $x$ is in degrees. - $\phi$ is the phase shift, here given as $30^\circ$. 3. **Finding amplitude $A$:** From the graph, the maximum value of $y$ is approximately 2 and the minimum is -2. Amplitude is half the distance between max and min: $$A = 2$$ 4. **Finding angular frequency $\omega$:** The wave completes one full cycle from $-90^\circ$ to $90^\circ$, so the period $T$ is: $$T = 90^\circ - (-90^\circ) = 180^\circ$$ Using the period formula: $$T = \frac{360^\circ}{\omega} \implies \omega = \frac{360^\circ}{T} = \frac{360^\circ}{180^\circ} = 2$$ 5. **Final answer:** $$A = 2, \quad \omega = 2$$ These values satisfy the given graph characteristics.