5. (5 marks) Find \( \frac{d^{99}}{d x^{99}}[\sin x] \), i.e. the \( 99^{\text {th }} \) derivative of \( \sin x \).

To find the \(99^{\text{th}}\) derivative of \(\sin x\), we first observe the pattern that emerges from taking successive derivatives of \(\sin x\): 1. \( \frac{d}{dx}[\sin x] = \cos x \) 2. \( \frac{d^2}{dx^2}[\sin x] = -\sin x \) 3. \( \frac{d^3}{dx^3}[\sin x] = -\cos x \) 4. \( \frac{d^4}{dx^4}[\sin x] = \sin x \) Notice that every four derivatives, the function returns to its original form. Thus, the derivatives of \(\sin x\) cycle every four: - \( \frac{d^0}{dx^0}[\sin x] = \sin x \) - \( \frac{d^1}{dx^1}[\sin x] = \cos x \) - \( \frac{d^2}{dx^2}[\sin x] = -\sin x \) - \( \frac{d^3}{dx^3}[\sin x] = -\cos x \) To find the \(99^{\text{th}}\) derivative, we calculate \(99 \mod 4\): \[ 99 \div 4 = 24 \quad \text{remainder } 3 \] This means \(99 \equiv 3 \mod 4\). According to the pattern we identified, \(\frac{d^3}{dx^3}[\sin x] = -\cos x\). Therefore, the \(99^{\text{th}}\) derivative of \(\sin x\) is: \[ \frac{d^{99}}{dx^{99}}[\sin x] = -\cos x \]