1. The problem is to show that $$\sinh 2x = 2 \sinh x \cosh x$$. 2. Recall the definitions of hyperbolic sine and cosine: $$\sinh x = \frac{e^x - e^{-x}}{2}$$ $$\cosh x = \frac{e^x + e^{-x}}{2}$$ 3. Using the definition of hyperbolic sine for $$2x$$: $$\sinh 2x = \frac{e^{2x} - e^{-2x}}{2}$$ 4. Now, calculate the right side: $$2 \sinh x \cosh x = 2 \times \frac{e^x - e^{-x}}{2} \times \frac{e^x + e^{-x}}{2} = \frac{(e^x - e^{-x})(e^x + e^{-x})}{2}$$ 5. Multiply the terms in the numerator: $$(e^x - e^{-x})(e^x + e^{-x}) = e^{2x} + e^x e^{-x} - e^{-x} e^x - e^{-2x} = e^{2x} - e^{-2x}$$ 6. Substitute back: $$2 \sinh x \cosh x = \frac{e^{2x} - e^{-2x}}{2}$$ 7. This matches exactly the expression for $$\sinh 2x$$ from step 3. Therefore, we have shown that: $$\sinh 2x = 2 \sinh x \cosh x$$