Question:

Step 1: Determine the range of $\theta$ in standard form. We need to simplify the interval $\theta \in \left[\dfrac{199\pi}{2}, 100\pi\right]$. $$\frac{199\pi}{2} = 99.5\pi = 49 \cdot 2\pi + 1.5\pi = 49 \cdot 2\pi + \frac{3\pi}{2}$$ $$100\pi = 50 \cdot 2\pi$$ Therefore, modulo $2\pi$, the interval corresponds to $\theta \in \left[\frac{3\pi}{2}, 2\pi\right]$, which is the **fourth quadrant**. Step 2: Determine the signs of $\sin\theta$ and $\cos\theta$ in the fourth quadrant. In the fourth quadrant where $\theta \in \left[\frac{3\pi}{2}, 2\pi\right]$: - $\sin\theta \leq 0$ (sine is negative or zero) - $\cos\theta \geq 0$ (cosine is positive or zero) Step 3: Express $\alpha$ using the sign of $\sin\theta$. Since $\sin\theta \leq 0$ in the fourth quadrant, we have $|\sin\theta| = -\sin\theta$. $$\alpha = \sin\theta|\sin\theta| = \sin\theta \cdot (-\sin\theta) = -\sin^2\theta$$ Step 4: Express $\beta$ using the sign of $\cos\theta$. Since $\cos\theta \geq 0$ in the fourth quadrant, we have $|\cos\theta| = \cos\theta$. $$\beta = \cos\theta|\cos\theta| = \cos\theta \cdot \cos\theta = \cos^2\theta$$ Step 5: Calculate $\alpha - \beta$. $$\alpha - \beta = -\sin^2\theta - \cos^2\theta = -(\sin^2\theta + \cos^2\theta)$$ Using the fundamental trigonometric identity $\sin^2\theta + \cos^2\theta = 1$: $$\alpha - \beta = -1$$ Step 6: Verify this is constant for all $\theta$ in the given interval. Since $\sin^2\theta + \cos^2\theta = 1$ is a fundamental identity that holds for all values of $\theta$, the result $\alpha - \beta = -1$ is constant throughout the entire interval $\left[\frac{199\pi}{2}, 100\pi\right]$. **Final Answer:** The correct option is **Option 4: $\alpha - \beta = -1$**