1. We are asked to analyze the function $$\sin\left(\frac{3\pi}{2} + x\right)$$. 2. Recall the sine addition formula: $$\sin(a + b) = \sin a \cos b + \cos a \sin b$$. 3. Apply the formula with $$a = \frac{3\pi}{2}$$ and $$b = x$$: $$\sin\left(\frac{3\pi}{2} + x\right) = \sin\frac{3\pi}{2} \cos x + \cos\frac{3\pi}{2} \sin x$$. 4. Evaluate sine and cosine at $$\frac{3\pi}{2}$$: $$\sin\frac{3\pi}{2} = -1$$ and $$\cos\frac{3\pi}{2} = 0$$. 5. Substitute these values: $$\sin\left(\frac{3\pi}{2} + x\right) = (-1) \cdot \cos x + 0 \cdot \sin x = -\cos x$$. 6. Therefore, the function simplifies to: $$\sin\left(\frac{3\pi}{2} + x\right) = -\cos x$$. 7. This means the original sine function shifted by $$\frac{3\pi}{2}$$ equals the negative cosine function. Final answer: $$\sin\left(\frac{3\pi}{2} + x\right) = -\cos x$$