Show that sinh2x equivalent to 2 coshxsinhx

The definitions are $\cosh x = \frac{e^x + e^{-x}}{2}$ and $\sinh x = \frac{e^x - e^{-x}}{2}$.

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

$$\text{RHS} = 2 \frac{(e^x + e^{-x})(e^x - e^{-x})}{4}$$ $$\text{RHS} = \frac{(e^x + e^{-x})(e^x - e^{-x})}{2}$$

Using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$:

$$\text{RHS} = \frac{(e^x)^2 - (e^{-x})^2}{2}$$ $$\text{RHS} = \frac{e^{2x} - e^{-2x}}{2}$$

The definition of $\sinh u$ is $\frac{e^u - e^{-u}}{2}$. With $u = 2x$:

$$\text{RHS} = \sinh(2x)$$

Since RHS = LHS, the identity is proven.

The final answer is $\boxed{\text{Shown}}$.
3 de faites, encore 2 aujourd'hui. Tu progresses.