When can sinh(x) and cosh(x) be equal?

Question

I know that for large positive numbers cosh(x) and sinh(x) would almost be equal to $e^x/2$ as $e^{-x}/2$ would become negligible given the magnitude of x in both cases. And so for a number like 31427.7920639882, sinh(x) and cosh(x) are equal. Apart from numbers being large, are there any other conditions at which sinh and cosh would be equal?

Answer 1

$$\cosh^2(x)=\sinh^2(x)+1$$

If they are equal, you get $$\cosh^2(x)=\cosh^2(x)+1$$

Answer 2

Never. Suppose $\sinh(x)=\cosh(x)$. Then, by definition, $$\frac{e^{x}+e^{-x}}{2}=\frac{e^{x}-e^{-x}}{2} \implies e^{-x}=0$$ So they get close together as $x \to \infty$, but they are never equal.

Answer 3

$sinh(x)=cosh(x)$ is equal to $e^{-x}=-e^{-x}$ which is: $2e^{-x}=0$, hence the answer is- there are never equal.

Your given number $31427.7920639882$ is actually not good example, may be you just calculated on computer, which have only approximations not the real values.

Answer 4

$$cosh(x)= \frac{e^x+e^{-x}}{2}$$ $$sinh(x)= \frac{e^x-e^{-x}}{2}$$ $$sinh(x)=cosh(x) \rightarrow \frac{e^x-e^{-x}}{2} = \frac{e^x+e^{-x}}{2}$$ $$e^x-e^{-x}= e^x+e^{-x}$$ $$e^x-e^{x}= e^{-x}+e^{-x}$$ $$0= 2e^{-x}$$ $$0= e^{-x}$$ $$ln(0_+)= -x$$ $$- \infty = -x$$ $$ \infty = x$$ These two functions only asymptotically approach each other as x$\rightarrow \infty$.