Proof that $-\sqrt{1+\sinh^2 (x)}=-\cosh⁡(x)$

Question

I am writing a paper in school and I am struggling to solve this equation. The starting equation is $-\sqrt{1+\sinh^2 (x)}=-\cosh⁡(x)$

And I just plugged in the definitions of cosh and sinh and came up with $-\sqrt{1 + \frac{e^x + e^{-x}}{2}}= -(\frac{e^x + e^{-x}}{2})$

Then I simplified and got the equation:

$\frac{(e^{2x + 1})^2\times e^{-2x}}{4}=\frac{e^{2x+1}\times e^{-x}}{2}$

But somehow this doesn't quite work out and I can't figure out the mistake.

Hint: prove $\cosh x\pm\sinh x=e^{\pm x}$, multiply, then use $\cosh x\ge1$. — J.G., Feb 05 '23 at 17:46
You have forgotten to square the $\sinh{x}$ when you rewrite it in exponential form. — Tanamas, Feb 06 '23 at 14:41

Yajat Shamji · Answer 1 · 2023-02-10T12:09:16.570

My proof of $-\sqrt{1 + \sinh^2x}=-\cosh x$ (using the exponential definitions and the equation given):

Divide both sides by $-1$: $\sqrt{1 + \sinh^2x}=\cosh x$

Substituting the definition of $\sinh x$ into the LHS gives $\sqrt{1 + \frac{(e^x - e^{-x})^2}{4}}$

Simplifying the expression in the square root gives $\sqrt{\frac{(e^x + e^{-x})^2}{4}}$

Taking the square root gives $\frac{e^x + e^{-x}}{2}$

Knowing that the definition of $\cosh x$ is $\frac{e^x + e^{-x}}{2}$, the LHS is equal to the RHS

Therefore, $-\sqrt{1 + \sinh^2x}\equiv-\cosh x$

What your mistake was you forgot to square the exponential form of $\sinh x$ in line $2$.

mjw · Answer 2 · 2023-02-06T15:07:45.227

0

Start with the definitions of $\cosh x$ and $\sinh x$.

Then show: $$\cosh^2 x - \sinh^2 x =1.$$

This leads to $$\cosh^2 x = 1+\sinh^2 x. \quad (*)$$

The hyperbolic cosine is always positive: $$\cosh x = \frac{e^x+e^{-x}}{2}>0.$$

Taking the positive square root of each side of (*):

$$\cosh x = \sqrt{1+\sinh^2 x}.$$

Multiply each side by $-1$.

edited Feb 06 '23 at 15:07

answered Feb 05 '23 at 18:16

mjw

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Important to note that the positive root is taken because $\cosh$ is always positive – preferred_anon Feb 05 '23 at 18:26
Yes, important detail. Thank you! – mjw Feb 05 '23 at 18:30
It should be $1+ \sinh^2x$. Not a minus sign. – peek-a-boo Feb 06 '23 at 14:18
Yes, Thank you! – mjw Feb 06 '23 at 15:02
Will update with corrections! – mjw Feb 06 '23 at 15:02

score 0 · Answer 3 · answered Feb 06 '23 at 14:19

Notice that $(e^x\pm e^{-x})^2=e^{2x}\pm2e^xe^{-x}+e^{-2x}=e^{2x}\pm2+e^{-2x}$. From this you get \begin{align*} \cosh^2x&=\frac{(e^x+e^{-x})^2}4=\frac{e^{2x}+2+e^{-2x}}4\\ \sinh^2x&=\frac{(e^x-e^{-x})^2}4=\frac{e^{2x}-2+e^{-2x}}4 \end{align*} and by subtracting these two equations you get $$\cosh^2x-\sinh^2x=1.$$ This means that $\sqrt{1+\sinh^2x}=\sqrt{\cosh^2x}=\cosh x$ (since $\cosh x>0$).

Proof that $-\sqrt{1+\sinh^2 (x)}=-\cosh⁡(x)$

3 Answers3