Question

There are some Formulas related to Hyperbolic Functions and Inverse Hyperbolic Functions:

Answer 1

\[\large \color{green}{\textbf{HYPERBOLIC FUNCTIONS:}}\]

\[\large 1.\quad sinhx = \frac{e^x-e^{-x}}{2}\]

\[\large 2. \quad coshx = \frac{e^x + e^{-x}}{2}\]

\[\large 3. \quad tanhx = \frac{e^x - e^{-x}}{e^x + e^{-x}}\]

\[\large 4. \quad cothx = \frac{e^x+ e^{-x}}{e^x - e^{-x}}\]

\[\large 5. \quad sechx = \frac{2}{(e^x + e^{-x})}\]

\[\large 6. \quad cosechx = \frac{2}{(e^x -e^{-x})}\]


\[\large \color{green}{\textbf{RELATIONS:}}\]

\[\large 1. \quad tanhx = \frac{sinhx}{coshx}\]

\[\large 2. \quad cothx = \frac{coshx}{sinhx}\]

\[\large 3. \quad sechx = \frac{1}{coshx}\]

\[\large 4. \quad cosechx = \frac{1}{sinhx}\]

\[\large 5. \quad \cosh^2x - \sinh^2x = 1\]

\[\large 6. \quad 1 - \tanh^2x = sech^2x\]

\[\large 7. \quad \coth^2x - 1 = cosech^2x\]



\[\large \color{green}{\textbf{NEGATIVE ANGLE PROPERTIES:}}\]

\[\large 1. \quad \sinh(-x) = -sinhx\]

\[\large 2. \quad \cosh(-x) = coshx\]

\[\large 3. \quad \tanh(-x) = -tanhx\]



\[\large \color{green}{\textbf{DOUBLE AND TRIPLE ANGLES FORMULAS:}}\]

\[\large 1. \quad \sinh2x = 2sinhx.coshx\]

\[\large 2. \quad \cosh2x = \cosh^2x + \sinh^2x\]

\[\large 3. \quad \tanh2x = \frac{2tanhx}{1 + \tanh^2 x}\]

\[\large 4. \quad \sinh3x = 3sinhx + 4\sinh^3x\]

\[\large 5. \quad \cosh3x = 4\sinh^3x -3coshx\]

\[\large 6. \quad \tanh3x = \frac{3tanhx + \tanh^3x}{1 + 3\tanh^2x}\]



\[\large \color{green}{\textbf{SUM AND DIFFERENCE OF ANGLES:}}\]

\[\large 1. \quad \sinh(\alpha \pm \beta) = \sinh \alpha.\cosh \beta \pm \cosh \alpha.\sinh \beta\]

\[\large 2. \quad \cosh(\alpha \pm \beta) = \cosh \alpha.\cosh \beta \pm \sinh \alpha.\sinh \beta\]

\[\large 3. \quad \tanh(\alpha \pm \beta) = \frac{\tanh \alpha \pm \tanh \beta}{1 \pm \tanh \alpha.\tanh \beta}\]



\[\large \color{green}{\textbf{SIMPLE VERSUS HYPERBOLIC:}}\]

\[\large 1. \quad cosix = coshx\]

\[\large 2. \quad sinix = i.sinhx\]

\[\large 2. \quad tanix = i.tanhx\]



\[\large \color{green}{\textbf{INVERSE HYPERBOLIC FUNCTIONS:}}\]

\[\large 1. \quad \sinh^{-1}x = Log_e(x + \sqrt{x^2 + 1})\]

\[\large 2. \quad \cosh^{-1}x = Log_e(x + \sqrt{x^2 - 1})\]

\[\large 3. \quad \tanh^{-1}x = \frac{1}{2}Log_e (\frac{1+x}{1-x})\]

\[\large 4. \quad \coth^{-1}x = \frac{1}{2}Log_e (\frac{x+1}{x-1})\]

\[\large 5. \quad sech^{-1}x = Log_e (\frac{1\pm \sqrt{1-x^2}}{x})\]

\[\large 6. \quad cosech^{-1}x = Log_e (\frac{1+ \sqrt{1+x^2}}{x}), \quad x > 0\]

\[\large = Log_e (\frac{1- \sqrt{1+x^2}}{x}), \quad x < 0\]



\[\large \color{green}{\textbf{RELATIONS IN INVERSE HYPERBOLIC:}}\]

\[\large 1. \quad \sinh^{-1}x = \cosh^{-1} \sqrt{1 + x^2}\]

\[\large 2. \quad \cosh^{-1}x = \sinh^{-1} \sqrt{x^2 - 1}\]

Answer 2

Very useful thanks :)

Answer 3

Welcome @mathslover

Answer 4

\[\huge \color{purple}{\textbf{<tips hat>}}\]

Answer 5

I have seen you every time you write it @lgbasallote
What it means??

Answer 6

http://en.wikipedia.org/wiki/Hat_tip

Answer 7

Ha ha ha ha..

Answer 8

it's an act i've been doing for 7 months now as a show of respect to people :p

Answer 9

Oh I got it now..
I have seen it in the picture in the link..
Ha ha ha..

Answer 10

What are \(\mathsf{\text{Hyperbolic Functions}}\)?

Answer 11

Do you know about simple Trigonometric Functions??

Also called Circular Trigonometric Functions...

Answer 12

I appreciate this @waterineyes , but could please help me understand what hyperbolic functions are actually?

Answer 13

These are just same as Normal or Circular Trigonometric Functions,
But these are expressed in exponential terms..
You can they are analogues of trigonometric functions..

Answer 14

Thanks, just one small example? Or may be a link where that explains this a bit lucidly? Thanks a bunch!

Answer 15

I have one pdf with me may be it will help you:

Answer 16

Thank you!!!

Answer 17

Welcome dear..
If you have any doubt then let me know that..

Answer 18

@waterineyes
can you please explain these a bit:
\[ \color{green}{\textbf{SIMPLE VERSUS HYPERBOLIC:}}\]\[1. \quad \cos ix = \cosh x\]\[2. \quad \sin ix = i\cdot\sinh x\]\[ 2. \quad \tan ix = i\cdot\tanh x\]

Answer 19

If you know Euler's Identities then you can easily verify these:

EULER IDENTITIES:

\[\large e^{i \theta} = \cos \theta + i.\sin \theta\]

\[\large e^{-i \theta} = \cos \theta - i.\sin \theta\]

Adding these both the identities:

\[\large \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2}\]

\[\large \sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i}\]

These are the two identities which will prove your question @UnkleRhaukus ,

Replace \(\theta\) by \(ix\) you get:

\[\large sinix = \frac{e^{i(ix) - e^{-i(ix)}}}{2i} \implies sinix = \frac{e^{-x} - e^x}{2i} \implies sinix = -\frac{e^x - e^{-x}}{2i}\]

\[\large \because \frac{-1}{i} = i\]
\[\large \therefore \quad sinix = i (\frac{e^x - e^{-x}}{2}) \implies sinix = i.sinhx\]

Similarly you can use this for proving \(cosix = coshx\)..

Getting @UnkleRhaukus ??

Answer 20

very nice

Answer 21

Thanks Uncle Rocks..