1. **Problem statement:** We need to express $\sin\theta - \sqrt{3} \cos\theta$ in the form $R \sin(\theta - \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. Then find the values of $\theta$ for which the expression attains its minimum value. 2. **Formula and rules:** The expression $a \sin\theta + b \cos\theta$ can be rewritten as $R \sin(\theta + \phi)$ where $R = \sqrt{a^2 + b^2}$ and $\tan\phi = \frac{b}{a}$. Here, since the form is $R \sin(\theta - \alpha)$, we use the identity: $$R \sin(\theta - \alpha) = R(\sin\theta \cos\alpha - \cos\theta \sin\alpha)$$ Matching coefficients: $$\sin\theta - \sqrt{3} \cos\theta = R \sin\theta \cos\alpha - R \cos\theta \sin\alpha$$ So, $$R \cos\alpha = 1$$ $$R \sin\alpha = \sqrt{3}$$ 3. **Find $R$ and $\alpha$:** Calculate $R$: $$R = \sqrt{(R \cos\alpha)^2 + (R \sin\alpha)^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$$ Calculate $\alpha$: $$\tan\alpha = \frac{R \sin\alpha}{R \cos\alpha} = \frac{\sqrt{3}}{1} = \sqrt{3}$$ Since $0 < \alpha < \frac{\pi}{2}$, we have: $$\alpha = \frac{\pi}{3}$$ 4. **Rewrite the expression:** $$\sin\theta - \sqrt{3} \cos\theta = 2 \sin\left(\theta - \frac{\pi}{3}\right)$$ 5. **Find minimum value and $\theta$ values:** The minimum value of $2 \sin(\theta - \frac{\pi}{3})$ is $-2$ because sine ranges from $-1$ to $1$. Set: $$\sin\left(\theta - \frac{\pi}{3}\right) = -1$$ This happens when: $$\theta - \frac{\pi}{3} = \frac{3\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}$$ So, $$\theta = \frac{\pi}{3} + \frac{3\pi}{2} + 2k\pi = \frac{11\pi}{6} + 2k\pi$$ **Final answer:** $$\sin\theta - \sqrt{3} \cos\theta = 2 \sin\left(\theta - \frac{\pi}{3}\right)$$ Minimum value is $-2$ at: $$\theta = \frac{11\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$$