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Prove the hyperbolic identities

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#1
Lim

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How to prove that ,


Without using the definitions of these hyperbolic functions.

#2
nemo

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I don't understand, if prove it without using the definitions of those functions then where we must start when any conceptions of maths were defined (!?)
<span style='color:purple'>Cây nghiêng không sợ chết đứng !</span>

#3
Lim

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I don't understand, if prove it without using the definitions of those functions then where we must start when any conceptions of maths were defined (!?)

Sorry if I made you confused .

I meant , is there any way to prove that identity rather than substitute the numbers in, like

. Because when people first studied hyperbolic functions, that identity could not come directly . They must start from something, and came to this result .

I want to know, where did they start and how they got to this result . Did they start from the trigonometric function , because of the analogy of sin function (sin(a+b) = sinacosb + sinbcosa ) or something else ?

What I'm looking is a geometric proof .
You have any suggestion ?

#4
bchl85

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Yeah, I think we can use sin(a+b) identity to prove it.
As we know sinix = isinhx and cosix = coshx
Thus isinh(x+y) = sin[i(x+y)] = sin(ix + iy) = sin(ix)cos(iy) + sin(iy)cos(ix)
=> isinh(x+y) = isinhx.coshy + isinhy.coshx
=> QED

Ah, there is a problem with your question ... where's "y" gone?

#5
Lim

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Yeah, I think we can use sin(a+b) identity to prove it.
As we know sinix = isinhx and cosix = coshx
Thus isinh(x+y) = sin[i(x+y)] = sin(ix + iy) = sin(ix)cos(iy) + sin(iy)cos(ix)
=> isinh(x+y) = isinhx.coshy + isinhy.coshx
=> QED

Ah, there is a problem with your question ... where's "y" gone?

Hey,
How do you get "sinix = isinhx and cosix = coshx"" ?

#6
bchl85

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Actually it's from the definition of http://dientuvietnam...metex.cgi?sinhx and http://dientuvietnam...etex.cgi?e^{ix}
You know
and you will have the formulae I used
I assumed that you already knew those formulae, sorry.
Ofcourse to do this question, using sin(a+b) you must start from that, it's not a start from definition of sinhx, but from material that you can use since it has been proved.




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