When proving hyperbolic identities why do we add one or two

Question

Example:

$\cosh^2x+\sinh^2x=\cosh2x$

Proof: $$\frac{1}{2}(e^x+e^{-x})^2+\frac{1}{2}(e^x-e^{-x})^2$$ Where does this two's come from? $$\frac{1}{4}(e^{2x}+e^{-2x}+2)+\frac{1}{4}(e^{2x}+e^{-2x}-2)$$

Answer 1

The two's come from expanding the brackets, i.e $(x+y)^2=x^2+y^2+\color{red}{2}xy$.

Since by definition we have

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

Then

$$\cosh^2(x)+\sinh^2(x)=\big(\frac{e^x+e^{-x}}{2}\big)^2+\big(\frac{e^x-e^{-x}}{2}\big)^2$$ $$ =\frac{1}{4}(e^x+e^{-x})^2+\frac{1}{4}(e^x-e^{-x})^2$$ $$=\frac{1}{4}(e^{x+x}+e^{-x-x}+2e^{x-x})+\frac{1}{4}(e^{x+x}+e^{-x-x}-2e^{x-x})$$ $$=\frac{1}{4}(e^{2x}+e^{-2x}+2)+\frac{1}{4}(e^{2x}+e^{-2x}-2)$$ $$=\frac{1}{4}(2e^{2x}+2e^{-2x})$$ $$=\frac{e^{2x}+e^{-2x}}{2}=\cosh(2x)$$