Question:

Step 1: Compute $x + y$ and $x - y$. $$x + y = \sin\theta\cos^3\theta + \sin^3\theta\cos\theta = \sin\theta\cos\theta(\cos^2\theta + \sin^2\theta) = \sin\theta\cos\theta = \frac{1}{2}\sin 2\theta$$ $$x - y = \sin\theta\cos^3\theta - \sin^3\theta\cos\theta = \sin\theta\cos\theta(\cos^2\theta - \sin^2\theta) = \frac{1}{2}\sin\theta\cos\theta \cdot 2\cos 2\theta = \frac{1}{4}\sin 4\theta$$ Step 2: Check option (a): $0 < \theta < \pi/4 \Rightarrow 0 < 2\theta < \pi/2 \Rightarrow \sin 2\theta > 0 \Rightarrow x + y > 0$. ✓ Step 3: Check option (b): $\pi/4 < \theta < \pi/2 \Rightarrow \pi < 4\theta < 2\pi$. For $x < y$, we need $x - y < 0$, i.e., $\sin 4\theta < 0$. For $\pi/4 < \theta < \pi/2$: $\pi < 4\theta < 2\pi$, so $\sin 4\theta$ can be positive or negative. Not always $< 0$. ✗ Step 4: Check option (c): $\pi/2 < \theta < 3\pi/4 \Rightarrow \pi < 2\theta < 3\pi/2 \Rightarrow \sin 2\theta < 0 \Rightarrow x + y < 0$. ✗ Step 5: Check option (d): $3\pi/4 < \theta < \pi \Rightarrow 3\pi < 4\theta < 4\pi$. For $x < y$: need $\sin 4\theta < 0$. For $3\pi < 4\theta < 4\pi$: $\sin 4\theta < 0$ for $3\pi < 4\theta < 4\pi$ (since $\sin$ is negative in $(3\pi, 4\pi)$ for the relevant range). Specifically $\sin 4\theta < 0$ for $3\pi < 4\theta < 4\pi$, so $x - y < 0 \Rightarrow x < y$. ✓