Geometric proof of $\cosh^2(x)-\sinh^2(x)=1$?

Asked Mar 12 '24 at 15:35

Active Mar 12 '24 at 15:35

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It is easy to show using the analytic continuation of $\sinh^2(x)$ and $\cosh^2(x)$ that the identity$$\cosh^2(x)-\sinh^2(x)=1$$holds. However, what I want to know is, is there any way to prove this geometrically?

I have been trying to prove it geometrically for around 3 or 4 hours, but I’m stuck on how I should start the proof in the first place, since all I’m really familiar with is the analytic proof of the identity.

asked Mar 12 '24 at 15:35

CrSb0001

2,532

What do $\sinh$ and $\cosh$ mean to you, geometrically? What approaches have you tried? – Arthur Mar 12 '24 at 15:41
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This is answered here: https://math.stackexchange.com/questions/526087/geometric-interpretation-of-hyperbolic-functions – Benjamin Dickman Mar 12 '24 at 15:42
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... and even more to the point, here: https://math.stackexchange.com/questions/757091/alternative-definition-of-hyperbolic-cosine-without-relying-on-exponential-funct/757241#757241 – Benjamin Dickman Mar 12 '24 at 15:43
@BenjaminDickman Ah, I did not notice that. Whoops – CrSb0001 Mar 12 '24 at 15:46
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@CrSb0001 all good, those two answers are also distributed across different questions. Hopefully you have a satisfactory answer now! – Benjamin Dickman Mar 12 '24 at 15:58

Geometric proof of $\cosh^2(x)-\sinh^2(x)=1$?

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