# Trigonometric - hyperbolic

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• Dec 26th 2010, 11:41 PM
trojsi
Trigonometric - hyperbolic
Hi guys,
just to make sure on the method i used on solving the identity attached.Attachment 20234

I first expanded the squared trigonometric functions and tangent, then substituted with j(phi) - converted to cosh and sinh etc. ending with :$\displaystyle 1-1-sinh^2\phi /cosh^2\phi =-sinh^2\phi /cosh^2\phi$
• Dec 27th 2010, 02:01 AM
CaptainBlack
Quote:

Originally Posted by trojsi
Hi guys,
just to make sure on the method i used on solving the identity attached.Attachment 20234

I first expanded the squared trigonometric functions and tangent, then substituted with j(phi) - converted to cosh and sinh etc. ending with :$\displaystyle 1-1-sinh^2\phi /cosh^2\phi =-sinh^2\phi /cosh^2\phi$

1. Sort out your brackets!

2. $\displaystyle \cos(2 i \phi)=\cosh(2\phi)$

CB
• Dec 27th 2010, 02:09 AM
Sudharaka
Quote:

Originally Posted by trojsi
Hi guys,
just to make sure on the method i used on solving the identity attached.Attachment 20234

I first expanded the squared trigonometric functions and tangent, then substituted with j(phi) - converted to cosh and sinh etc. ending with :$\displaystyle 1-1-sinh^2\phi /cosh^2\phi =-sinh^2\phi /cosh^2\phi$

Dear trojsi,

You do not need to expand the trignometric functions. Replace $\displaystyle \phi~with~i\phi$ and use, $\displaystyle \cos(i\phi)=\cosh \phi~and~\tan(i\phi)=i\tanh \phi$

The correct identity is, $\displaystyle -\tanh^2\phi=1-\dfrac{\cosh 2\phi}{\cosh^2\phi}$

Hope this helps you.
• Dec 27th 2010, 02:23 AM
trojsi
Thanks a lot, I manage to get it with those identities.