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April 19th 2008, 02:52 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 $\mathrm{cosh} 2x = 2 \mathrm{cosh}2x - 1$ .

Thanks in advance. (Smile)
April 19th 2008, 03:06 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 $\mathrm{cosh} 2x = 2 \mathrm{cosh}2x - 1$ .

Thanks in advance. (Smile)

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

$\cosh (2x) = \frac{e^{2x} + e^{-2x}}{2} = \frac{(e^x + e^{-x})^2 - 2}{2} = \frac{(e^x + e^{-x})^2}{2} - \frac{2}{2} = 2 \left(\frac{e^x + e^{-x}}{2}\right)^2 - 1$ .......
April 19th 2008, 03:24 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 $-2$ ?
April 19th 2008, 03:34 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 $e^{2x} + e^{-2x} = (e^{x} + e^{-x})^2 - 2$ .

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