# Proving...

• Apr 19th 2008, 02:52 AM
Simplicity
Proving...
Q: Using the definitions of $\displaystyle \mathrm{cosh} x$ and $\displaystyle \mathrm{sinh} x$ in terms of $\displaystyle e^x$ and $\displaystyle e^{-x}$, prove that $\displaystyle \mathrm{cosh} 2x = 2 \mathrm{cosh}2x - 1$.

• Apr 19th 2008, 03:06 AM
mr fantastic
Quote:

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

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

$\displaystyle \cosh (2x) = \frac{e^{2x} + e^{-2x}}{2} = \frac{(e^x + e^{-x})^2 - 2}{2} = \frac{(e^x + e^{-x})^2}{2} - \frac{2}{2} = 2 \left(\frac{e^x + e^{-x}}{2}\right)^2 - 1$ .......
• Apr 19th 2008, 03:24 AM
Simplicity
Quote:

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

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

How did you get $\displaystyle -2$?
• Apr 19th 2008, 03:34 AM
mr fantastic
Quote:

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

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

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