showing the identity of a hyperbolic function

Question

Need to just verify that I'm doing this one right.

1. $$\sinh{2x} = 2 \cdot \sinh{x} \cdot \cosh{x}$$ $$= 2 \cdot \frac{e^x - e^{-x}}{2} \cdot \frac{e^x + e^{-x}}{2}$$ $$= \frac{2 \cdot e^{2x} - e^{-2x}}{4}$$

$$= \sinh{2x}$$

Is that right?

Answer 1

Your idea is good, but it's not laid out well.

An identity can be verified by transforming one side into the other or both into the same expression. In this case it's easier starting from the right-hand side, which is essentially what you do: \begin{align} 2\sinh x\cosh x &=2\frac{e^x-e^{-x}}{2}\frac{e^x+e^{-x}}{2}\\[6px] &=\frac{(e^x-e^{-x})(e^x+e^{-x})}{2}\\[6px] &=\frac{e^{2x}-e^{-2x}}{2}\\[6px] &=\sinh2x \end{align} using $(e^x)^2=e^{2x}$ and $(e^{-x})^2=e^{-2x}$.