Hyperbolics - Use Osborne's Rule to show sech^2x=1-tanh^2x

Im not quite sure what I need to do,

I know

sech^2x = 1/cosh(x)

I know tanhx = sinhx/coshx

I know that in trig 1+tan^2x would be sec^2x

and I know it is 1-tanh^2x because of the (-) in the sinhx

But, I dont know how to put this all together to prove it?

Re: Hyperbolics - Use Osborne's Rule to show sech^2x=1-tanh^2x

Quote:

Originally Posted by

**tjnortham** Im not quite sure what I need to do,

I know

sech^2x = 1/cosh(x)----------should be sech(x)=1/cosh(x)

I know tanhx = sinhx/coshx

I know that in trig 1+tan^2x would be sec^2x

and I know it is 1-tanh^2x because of the (-) in the sinhx

But, I dont know how to put this all together to prove it?

I hope you know the definitions of hyperbolic functions.

$\displaystyle \cosh(x)=\frac{e^{-x}+e^x}{2}\mbox{ and }\sinh(x)=\frac{e^{-x}-e^x}{2}$

1) Find $\displaystyle \tanh(x)$ and hence $\displaystyle 1-\tanh^{2}(x)$.

2) Find $\displaystyle \text{sech}^{2}(x)=\frac{1}{\cosh^{2}(x)}$

Compare the two results.

Re: Hyperbolics - Use Osborne's Rule to show sech^2x=1-tanh^2x

Quote:

Originally Posted by

**tjnortham** Im not quite sure what I need to do,

I know

sech^2x = 1/cosh(x)

I know tanhx = sinhx/coshx

I know that in trig 1+tan^2x would be sec^2x

and I know it is 1-tanh^2x because of the (-) in the sinhx

But, I dont know how to put this all together to prove it?

$\displaystyle \displaystyle \begin{align*} \textrm{sech}\,{(x)} &= \frac{1}{\cosh{(x)}} \\ &= \frac{2}{e^x + e^{-x}} \\ \textrm{sech}^2{(x)} &= \left(\frac{2}{e^x + e^{-x}}\right)^2 \\ &=\frac{4}{e^{2x} + 2 + e^{-2x}} \end{align*}$

$\displaystyle \displaystyle \begin{align*} \tanh{(x)} &= \frac{\sinh{(x)}}{\cosh{(x)}} \\ &= \frac{\frac{e^{x} - e^{-x}}{2}}{\frac{e^x + e^{-x}}{2}} \\ &= \frac{e^x - e^{-x}}{e^x + e^{-x}} \\ \tanh^2{(x)} &= \left(\frac{e^x - e^{-x}}{e^x + e^{-x}}\right)^2 \\ &= \frac{e^{2x} - 2 + e^{-2x}}{e^{2x} + 2 + e^{-2x}} \\ 1 - \tanh^2{(x)} &= 1 - \frac{e^{2x} - 2 + e^{-2x}}{e^{2x} + 2 + e^{-2x}} \\ &= \frac{e^{2x} + 2 + e^{-2x}}{e^{2x} + 2 + e^{-2x}} - \frac{e^{2x} - 2 + e^{-2x}}{e^{2x} + 2 + e^{-2x}} \\ &= \frac{4}{e^{2x} + 2 + e^{-2x}} \\ &= \textrm{sech}^2{x} \end{align*}$

Re: Hyperbolics - Use Osborne's Rule to show sech^2x=1-tanh^2x

Quote:

Originally Posted by

**tjnortham** Im not quite sure what I need to do,

I know

sech^2x = 1/cosh(x)

I know tanhx = sinhx/coshx

I know that in trig 1+tan^2x would be sec^2x

and I know it is 1-tanh^2x because of the (-) in the sinhx

But, I dont know how to put this all together to prove it?

I haven't seen Osborne's rule used in any of the previous methods. At least, not directly. You know that $\displaystyle 1+tan^2(x)=sec^2(x)$

Osborne's rule simply states that you can convert trigonometric identities to hyperbolic identities, except where there is a product of sines, when you have to change the sign. You know that: you reference it in the op.

So, why not just prove $\displaystyle 1+tan^2(x)=sec^2(x)$ and change the signs where appropriate?