Prove that sinhx, i feel like im going round in circles

OP

thanks for the quick replies, i didnt find those formulas in my notes for some reasons, ill give it a go now i have those and report back if im still clueless haha

Reply 4

TheBBQ

21

When you have something like this, it's usually a good idea to try and break down the more complicated looking side.

Using the definition of tanh(x/2) i.e. it's exponential form will always get you there, but it can sometimes get quite cumbersome when multiplying and dividing through all the exponentials.

(edited 9 years ago)

Reply 5

OP

Hey again guys had another crack and Still can seem to work it out, can anyone gimmie a hand ?

Reply 6

there are 2 ways of doing this (variations)

method A

start from the right

write the tanh(x/2) in terms of sinh(x/2) and cosh(x/2)

Reply 7

OP

2tanhx/2=2sinhx/2coshx+1?

Reply 8

Original post by Ramjam

2tanhx/2=2sinhx/2coshx+1?

sorry no.

do you know the definition of the hyperbolic tangent?

if you do start at the RHS and look at the suggestion of my previous post

Reply 9

Original post by Ramjam

2tanhx/2=2sinhx/2coshx+1?

also use the reply button please if the message is for me

otherwise I do not get a notification so I can only see the reply when I refresh the screen.

Reply 10

OP

Original post by TeeEm

also use the reply button please if the message is for me

otherwise I do not get a notification so I can only see the reply when I refresh the screen.

Sorry bud thought I did press it, no I don't know the hyperbolic tangent

(edited 9 years ago)

Reply 11

Original post by Ramjam

Sorry bud thought I did press it, no I don't know the hyperbolic tangent

ok

tanh(x/2) = sinh(x/2) / cosh(x/2), in analogy with the trigonometric tangent.

start with the RHS

replace all tanh(x/2) with sinh(x/2) / cosh(x/2)

Reply 12

Phichi

11

Before changing tanh for cosh and sinh, I'd start with the most basic identity:

cosh^2x-sinh^2x=1

Diving through by

cosh^2x

would give you an equation of the form you could make use of. Then as TeeEm said, changing the

tanh(\frac{x}{2})

with both

sinh(\frac{x}{2})

and

cosh(\frac{x}{2})

would allow you to simplify, using what you found with the first identity.

Bare in mind, the double angle formulae you used for normal trig functions, are practically the same here, with the use of Osborne's rule, which won't occur here.

In trig,

Sin(2x)=2Cos(x)Sin(x)

and in hyperbolics

Sinh(2x)=2Cosh(2x)Sinh(2x)

. Making the simplification very easy. You'll end up with something that looks like the previous, if you simplify correctly, feel free to ask for help though.

(edited 9 years ago)

Reply 13

OP