Question

Can anyone show that " sinhx is (e^x-e^-x)/2 "

Answer 1

That's the definition of sinh(x).

Answer 2

I know ...I would like to know how it was derived :)

Answer 3

Some history of its discovery ...

Answer 4

Oh. It's an analog to sin(x), except where
\[ \cos^2(x)+\sin^2(x) = 1\]
which is characteristic of a circle,
\[ \cosh^2(x)-\sinh^2(x) = 1\]
which is characteristic of a hyperbola. The definitions you are familiar with satisfy that condition, and if you analogously define tanh(x) and the rest of the hyperbolic functions you find many of the trig identities are very similar to the hyperbolic ones. If you study complex analysis, you find that
\[ \sin(ix)=i\sinh(x) \]
\[\cos(ix)= \cosh(x) \]
which explains this.

Answer 5

There is a popular equation as \[e ^{i*\theta} = \cos(\theta) + i*\sin(\theta) \], the proof of which lies in the series expansion of these functions. Thus, we also have, \[e ^{-i*\theta} = \cos(-\theta) + i*\sin(-\theta) = \cos(\theta) - i*\sin(\theta) \] . Once one assumes this as true, we can use this to define \[sin(\theta) = ({e ^{i*\theta} - e ^{-i*\theta}})/2*i \]. On similar lines, people thought of defining a function as \[sinh(\theta) = ({e ^{\theta} - e ^{-\theta}})/2 \], and because it looks so similar to sin, they called it sin + hyperbolic, because its graph looks like a hyperbolic function

Answer 6

I was about to go to Euler's formula I just can't see why sinh i x = i sin x .... .

Answer 7

The hyperbolic analog of Euler's formula : e^x = cosh x + sinh x, so how was deduced that sinh x = -i sin i x ... boy, do I have to go back in time !

Answer 8

do you guys have a link where I can read about it ?