# Proving...

• Apr 19th 2008, 03:52 AM
Simplicity
Proving...
Q: Using the definitions of $\mathrm{cosh} x$ and $\mathrm{sinh} x$ in terms of $e^x$ and $e^{-x}$, prove that $\mathrm{cosh} 2x = 2 \mathrm{cosh}2x - 1$.

• Apr 19th 2008, 04:06 AM
mr fantastic
Quote:

Originally Posted by Air
Q: Using the definitions of $\mathrm{cosh} x$ and $\mathrm{sinh} x$ in terms of $e^x$ and $e^{-x}$, prove that $\mathrm{cosh} 2x = 2 \mathrm{cosh}2x - 1$.

Prove: $\mathrm{cosh} 2x = 2 \mathrm{cosh}^2 x - 1$.

$\cosh (2x) = \frac{e^{2x} + e^{-2x}}{2} = \frac{(e^x + e^{-x})^2 - 2}{2} = \frac{(e^x + e^{-x})^2}{2} - \frac{2}{2} = 2 \left(\frac{e^x + e^{-x}}{2}\right)^2 - 1$ .......
• Apr 19th 2008, 04:24 AM
Simplicity
Quote:

Originally Posted by mr fantastic
Prove: $\mathrm{cosh} 2x = 2 \mathrm{cosh}^2 x - 1$.

$\cosh (2x) = \frac{e^{2x} + e^{-2x}}{2} = \frac{(e^x + e^{-x})^2 \mathbf{- 2}}{2} = \frac{(e^x + e^{-x})^2}{2} - \frac{2}{2} = 2 \left(\frac{e^x + e^{-x}}{2}\right)^2 - 1$ .......

How did you get $-2$?
• Apr 19th 2008, 04:34 AM
mr fantastic
Quote:

Originally Posted by Air
How did you get $-2$?

$(e^{x} + e^{-x})^2 = e^{2x} + 2 (e^x)(e^{-x}) + e^{-2x} = e^{2x} + 2 + e^{-2x}$.

Therefore $e^{2x} + e^{-2x} = (e^{x} + e^{-x})^2 - 2$.