1. The problem is to analyze the function $y = \sin 5x \cdot \cos 3x$. 2. We can use the product-to-sum identity for sine and cosine: $$\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$$ 3. Applying this identity with $A = 5x$ and $B = 3x$, we get: $$y = \sin 5x \cos 3x = \frac{1}{2} [\sin(5x + 3x) + \sin(5x - 3x)] = \frac{1}{2} [\sin 8x + \sin 2x]$$ 4. This expression shows that $y$ is the sum of two sine functions with different frequencies. 5. The function has zeros where either $\sin 8x = -\sin 2x$ or both terms are zero. 6. The amplitude of $y$ varies between $-1$ and $1$ because sine functions have amplitude 1 and the factor $\frac{1}{2}$ scales the sum. Final answer: $$y = \frac{1}{2} (\sin 8x + \sin 2x)$$