# In proving $\cosh(\sinh^{-1}(x))=\sqrt{1+x^2}$ don't we need restrictions on values of $x$?

1. The functions $$\sinh$$ and $$\tanh$$ are one-one; their inverses $$\sinh^{-1}$$ and $$\tanh^{-1}$$ are defined on $$\mathbb{R}$$ and $$(-1,1)$$, respectively. These inverse functions are sometimes denoted by $$\arg\sinh$$ and $$\arg\tanh$$ (the "argument" of the hyperbolic sine and tangent. If $$\cosh$$ is restricted to $$[0,\infty)$$ it has an inverse, denoted by $$\arg\cosh$$, or simply $$\cosh^{-1}$$, which is defined on $$[1,\infty)$$.

Prove, using information from problem $$8$$, that

(b) $$\cosh(\sinh^{-1}(x))=\sqrt{1+x^2}\tag{1}$$

Shouldn't there be a restriction on the values that $$x$$ can take in $$(1)$$ (such restrictions are present in other items in this problem, for example)?

In problem $$8$$ we proved that

$$\cosh^2-\sinh^2=1\tag{1}$$

Hence, expression $$(1)$$ evaluated at a point $$\sinh^{-1}(x)$$ is

$$\cosh^2(\sinh^{-1}(x))-\sinh^2(\sinh^{-1}(x))=1$$

$$\cosh^2(\sinh^{-1}(x))-x^2=1$$

$$\cosh^2(\sinh^{-1}(x))=1+x^2\tag{2}$$

$$\cosh(\sinh^{-1}(x))=\sqrt{1+x^2}\tag{3}$$

Consider the step from $$(2)$$ to $$(3)$$.

In $$(2)$$ don't we need to restrict $$\sinh^{-1}(x)$$ to be in $$[1,\infty)$$?

Ie

$$\sinh^{-1}(x)\in [1,\infty)$$

Now, since $$\sinh$$ is increasing and

$$\sinh^{-1}(x)=1 \implies x=\sinh{1}$$

then if $$x\in[\sinh{1},\infty)$$ then $$\sinh^{-1}(x) \in [1,\infty)$$.

Finally, just to confirm, the reason we take the positive square root is because $$\cosh$$ always positive, correct?

• Your $\sin\,$s should be $\sinh\,$s.
– anon
Jul 27, 2022 at 5:49
• In the real variable we don't, because $\sinh:\Bbb R\to\Bbb R$ is bijective and $\cosh:\Bbb R\to [1,\infty)$ is non-negative. Jul 27, 2022 at 5:50
• The reason why we take the positive square root for $\cosh$ is partially that $\cosh\ge0$ and it's probably inherent to the proof you're reading, but it should be noted that $\sinh^{-1}x$ has the explicit formula $\ln\left(x+\sqrt{x^2+1}\right)$, so you could just compute $\cosh\sinh^{-1}(x)$ directly in terms of elementary functions. Jul 27, 2022 at 5:54
• In $\cosh(\text{arcsinh}(x))$, $\text{arcsinh}(x)$ is the output which is fed into the function $\cosh(x)$. The possible values of this output (the image) is the set of real numbers, which comes from the domain of the inverse function $\sinh(x) = \frac{e^x+e^{-x}}{2}$. Jul 27, 2022 at 5:56
• Hence there should be no restrictions on the domain of $\cosh(\text{\arcsinh}(x))$. Jul 27, 2022 at 5:57

The domain of $$\cosh^{-1}$$ may be $$[1,\infty)$$, but there is no $$\cosh^{-1}$$ in the identity you're talking about so this interval is irrelevant. The domain of $$\sinh^{-1} x$$ is all real numbers; its range is all real numbers; the domain of $$\cosh$$ is all real numbers; so $$\cosh(\sinh^{-1}x)$$ is well-defined for all $$x$$.

Shouldn't there be a restriction on the values that $$x$$ can take in $$\cosh(\sinh^{-1}(x))=\sqrt{1+x^2}$$ (such restrictions are present in other items in this problem, for example)?

No.

In $$\cosh^2(\sinh^{-1}(x))=1+x^2$$ don't we need to restrict $$\sinh^{−1}(x)$$ to be in $$[1,\infty)$$?

No.

Finally, just to confirm, the reason we take the positive square root is because cosh always positive, correct?

Yes.

The formula holds without any restrictions, since $$sinh(x)$$ is strictly increasing and hence invertible on $$\mathbb{R}$$. Let $$sinh^{-1}(x)=y$$. Then $$x=sinhy=\dfrac{e^{y}-e^{-y}}{2}$$ .

Soving the binomial for $$e^{y}$$ we get $$e^{y}=x\pm \sqrt{x^{2}+1}$$. Since $$e^{y}$$ is positive, the minus case is rejected and we obtain $$e^{y}=x+\sqrt{x^{2}+1}$$ and $$y=ln(x+\sqrt{x^{2}+1})$$.

Now $$cosh(y)$$=$$\dfrac{x+\sqrt{x^{2}+1}}{2}$$ +$$\dfrac{1}{2(x+\sqrt{x^{2}+1})}$$=$$\sqrt{x^{2}+1}$$.

Thus no restrictions on $$x$$ are required!!

• There are strictly increasing functions for which you still need to restrict the domain when you take sections, namely $\cosh(\exp^{-1}(x))=\cosh\ln x$ has domain $(0,\infty)$. Jul 27, 2022 at 13:21
• OK ! But this is a complete proof that there is no need for restrictions in this case!
– user1054388
Jul 27, 2022 at 13:28
• ... then why leave the false statements there? Jul 27, 2022 at 13:32
• Which are the "false statements"?? Please explain!
– user1054388
Jul 27, 2022 at 13:34
• "The formula holds without any restrictions, since sinh(x) is strictly increasing and hence invertible on R." Jul 27, 2022 at 13:34