Prove that sinh(2x) = 2sinh(x)cosh(x)

Question

When trying to solve a textbook question I'm getting a wrong result but I'm pretty sure I'm not doing anything wrong. Can somebody spot my mistake?

Question:

Prove that $$sinh(2x) = 2sinh(x)cosh(x)$$

My (wrong) solution:

$$sinh(x) = \frac{e^x - e^{-x}}{2}$$ $$sinh(2x) = \frac{e^{2x} - e^{-2x}}{2}$$ $$sinh(2x) = \frac{(e^{x} + e^{-x})(e^{x} - e^{-x})}{2}$$ $$sinh(2x) = cosh(x)sinh(x)$$

So I'm missing a multiplication by 2. What is my mistake?

Answer 1

$$\sinh2x=\frac{(e^x-e^{-x})(e^x+e^{-x})}{2}=2\cdot\frac{e^x-e^{-x}}{2}\cdot\frac{e^x+e^{-x}}{2}=2\sinh{x}\cosh{x}.$$