Proving...

**Simplicity** · April 19th 2008, 02:52 AM

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.

**mr fantastic** · April 19th 2008, 03:06 AM

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.

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

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

**Simplicity** · April 19th 2008, 03:24 AM

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 $-2$ ?

**mr fantastic** · April 19th 2008, 03:34 AM

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

Thread: Proving...

LinkBack

Thread Tools

Display

Proving...

Similar Math Help Forum Discussions

Proving?

Proving the SST = SSE + SSR

proving sin^2(x)<=|sin(x)|

proving

Proving an identity that's proving to be complex

Search Tags