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

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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?

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1 Answer

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

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