<s>[INST] Hyperbolic functions proof using $\cosh x$ 
Prove that $\cosh x-1\equiv \frac12(e^{0.5x}-e^{-0.5x})^2$

I'm stuck on what appears to be the last step, please could someone explain where I have made a mistake?
\begin{align}
\frac12(e^{0.5x}-e^{-0.5x})^2 & \equiv \frac12(\sinh(0.5x))^2 \\
& \equiv \frac12\sinh^2(0.5x) \\
& \equiv \frac12(\cosh^2(0.5x) -1) \\
& \equiv \frac12((\frac{e^{0.5x}+e^{-0.5x}}{2})^2 -1) \\ 
& \equiv \frac12((\frac{e^x+e^{-x}+2}4) -1) \\ 
& \equiv \frac14((\frac{e^x+e^{-x}+2}2)-2) \\ 
& \equiv \frac14(\frac{e^x+e^{-x}}2+\frac22-2) \\
& \equiv \frac14(\frac{e^x+e^{-x}}{2}-1) \\
& \equiv \frac14(\cosh x-1) \\
\end{align}
which is a quarter of what I am trying to prove?
 [/INST] Note that $$\begin{align}
\frac12(e^{0.5x}-e^{-0.5x})^2 & \equiv \frac12(2\sinh(0.5x))^2 \\
& \equiv 2\sinh^2(0.5x) \\
& \equiv 2(\cosh^2(0.5x) -1) \\
& \equiv 2((\frac{e^{0.5x}+e^{-0.5x}}{2})^2 -1) \\ 
& \equiv 2((\frac{e^x+e^{-x}+2}4) -1) \\ 
& \equiv ((\frac{e^x+e^{-x}+2}2)-2) \\ 
& \equiv (\frac{e^x+e^{-x}}2+\frac22-2) \\
& \equiv (\frac{e^x+e^{-x}}{2}-1) \\
& \equiv (\cosh x-1) \\
\end{align}$$ Which is what you wanted to prove.  
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