If sinh(y)=x, then, show that cosh(y)=sqrt(1+x^2)

[cks]

if sinh(y)=x, then, show that cosh(y)=sqrt(1+x^2)

I know how to prove but i have difficult in choosing the signs.

sinh(y) = x = [exp(y)-exp(-y)]/2

there are two equations I can find

exp(2y) - 2x exp(y) -1 =0 OR exp(-2y) + 2xexp(-y) -1 =0

exp(y) = x +- sqrt(x^2 +1) & exp(-y) = -x +- sqrt(x^2 + 1)

if I select the signs of +

exp(y) = x + sqrt(x^2 +1) & exp(-y) = -x + sqrt(x^2 + 1)

then by substituting to cosh(y) = [exp(y) + exp(-y)] / 2

then, I can find the answer of
cosh(y)=sqrt(1+x^2)

But, what should I say to justify my actions of selecting the +ve sign.

 

[neutrino]

Why not just substitute $$x = \frac{\left(e^y - e^{-y}\right)}{2}$$ into the expression $$\sqrt{1+x^2}$$ and simplify?

 

[cks]

Thanks, but I'd like to find cosh(y) without prior knowledge that it's equal to sqrt(1+x^2)

 

[robphy]

For real y, exp(y)>0.

 

[arildno]

Utilize the identity:
$$Cosh^{2}(y)-Sinh^{2}(y)=1$$
Along with the requirement that Cosh(y) is a strictly positive function.
(Alternatively, that it is an even function with Cosh(0)=1)

 

[cks]

still don't quite understand

 

[Gib Z]

arildno pretty much gave you the identity that you need. If you need to prove it, Just replace cosh y and sinh y with their exponential definitions and simplify. Then isolate cosh y on the left hand side and it is clear. Basically when we take the final square root, one could normally argue you could have either the positive or the negative root. However cosh is always positive.

 

[robphy]

There are different ways to approach the title of this thread, depending on what you are given to start with and what constraints are imposed... which should include your working definitions of sinh and cosh.

Rather than merely proving an identity, it appears that the OP wishes to obtain the title by using the definitions of sinh and cosh starting from their definitions using the sum and difference of exponential functions. In addition, one might taking the point of view that one doesn't know at this stage any properties of cosh except for its working definition. So, it might be "cheating" to use any other properties of cosh which you didn't derive already.

cks, is this correct?

(If I'm not mistaken, my suggestion "For real y, exp(y)>0" provides a reason to choose the positive sign.)

 

[cks]

Oh, I see thank you. Let me rephrase what you all said

exp(y) = x +- sqrt(x^2 +1) & exp(-y) = -x +- sqrt(x^2 + 1)

exp(y) is always > 0

so if we select exp(y) = x - sqrt(x^2 +1)

and most importantly , sqrt(x^2 +1) > x

so exp(y) < 0 which is false.

As a result, we have to select exp(y) = x + sqrt(x^2 +1)

The similar argument can be applied to exp(-y)

Thank you all of you, really appreciate that.