Answer: D
The minimum value of \(a\sin\theta + b\cos\theta\) is \(-\sqrt{a^2 + b^2}\), and the maximum value is \(\sqrt{a^2 + b^2}\). Comparing \(4\cos\theta - 3\sin\theta\) with the form \(a\sin\theta + b\cos\theta\), we get: \(a = -3\), \(b = 4\) \[ \therefore -\sqrt{a^2 + b^2} = -\sqrt{9 + 16} = -\sqrt{25} = -5 \] \[ \therefore \text{The minimum value of } 4\cos\theta - 3\sin\theta \text{ is } -5 \]
The minimum value of \(a\sin\theta + b\cos\theta\) is \(-\sqrt{a^2 + b^2}\), and the maximum value is \(\sqrt{a^2 + b^2}\). Comparing \(4\cos\theta - 3\sin\theta\) with the form \(a\sin\theta + b\cos\theta\), we get: \(a = -3\), \(b = 4\) \[ \therefore -\sqrt{a^2 + b^2} = -\sqrt{9 + 16} = -\sqrt{25} = -5 \] \[ \therefore \text{The minimum value of } 4\cos\theta - 3\sin\theta \text{ is } -5 \]