Prove arcsinh sinhx = x

...As the title says, using arcsinh(x) = ln (x +(y^2+1)^0.5)

I've substituted it all in, writing sinh in terms of e^x and stuff but I don't know where to go from there...

Any help would be much appreciated! :smile:

OP

Anyone?

http://www.thestudentroom.co.uk/showthread.php?p=34846553

Reply 3

username667907

4

i know this is UG stuff, and the only experience of this is have is FP2, but could you say:

arc\sinh {(\sinh x)} = \ln{\left| \frac{e^x - e^{-x}}{2} + \sqrt{( \frac{e^x - e^{-x}}{2} )^2 + 1}\right| }

then put it all in the square root and take out the power of 1/2:

arc\sinh {(\sinh x)} = \frac{1}{2} \ln{\left| 2(\frac{e^x - e^{-x}}{2})^2 + 1 \right| }

simplifying: (please correct my maths)

arc\sinh {(\sinh x)} = \frac{1}{2} \ln{\left| \frac{e^x - e^{-x}}{2} \right| }

ermm...
then i dont know

(edited 12 years ago)

Reply 4

mzwoteva16

OP

2

Original post by ghostwalker

http://www.thestudentroom.co.uk/showthread.php?p=34846553

I've done that now and got:

ln( 0.5 [ (e^2x) + e^-2x)^0.5 + (e^x) - (e^-x)]

Can I multiply out (e^2x) + e^-2x)^0.5 now?

Thank you btw!

(edited 12 years ago)

Reply 5

ghostwalker

17

Original post by mzwoteva16

I've done that now and got:

ln( 0.5 [ (e^2x) + e^-2x)^0.5 + (e^x) - (e^-x)]

Can I multiply out (e^2x) + e^-2x)^0.5 now?

Thank you btw!

You've made an error in getting to the first line there.

Have another go, or post your working.

Reply 6

k9markiii

18

Original post by mzwoteva16

...As the title says, using arcsinh(x) = ln (x +(y^2+1)^0.5)

I've substituted it all in, writing sinh in terms of e^x and stuff but I don't know where to go from there...

Any help would be much appreciated! :smile:

arcsinh(x)=y (sinh both sides)
x=sinh(y)

Identity 1:

sinh(y)+cosh(y)=e^y

(\frac{e^y-e^{-y}+e^y+e^{-y}}{2}=e^y)

Identity 2:

sinh^2(y)-cosh^2(y)=-1

(e^(2y)-2e^(0)+e^(-2y))/(4) -(e^(2y)+2e^0+e^(-2y))/4=-4/4)

latex is not working for me.

Rearrange identity 2 to find cosh(y) in terms of sin(y)

\sqrt{{sinh^2(y)+1}}=cosh^2(y)

Use identity 1:
sinh(y)+√sinh^2(y)+1)=e^y (ln both sides)
ln|sinh(y)+sinh^2(y)+1|=y=arcsin(x)

Reply 7

ztibor

10

Original post by Phredd

i know this is UG stuff, and the only experience of this is have is FP2, but could you say:

arc\sinh {(\sinh x)} = \ln{\left| \frac{e^x - e^{-x}}{2} + \sqrt{( \frac{e^x - e^{-x}}{2} )^2 + 1}\right| }

then put it all in the square root and take out the power of 1/2:

arc\sinh {(\sinh x)} = \frac{1}{2} \ln{\left| 2(\frac{e^x - e^{-x}}{2})^2 + 1 \right| }

You can not do it
Expand the square under the square root then arrange and you will get a complete square there. Taking the root and arranging you will get lne^x

Reply 8

mzwoteva16

OP

2

Thank you everyone, I managed it, I was just being slow and not realising that cosh^2(x) - sinh^2(x)= 1 thus ending up with a horrible algebraic mess. All done and dusted now! :smile:

Prove arcsinh sinhx = x

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