# Proving Hyperbolic Sine and Tan functions are bijective.

I'm working on this paper where we're supposed to get to know the hyperbolic trig functions... and we have to prove a bunch of things about them, like their derivatives and such. I have this question, which I think I have correctly proven, I'm just not sure if it's correct.

1) Show that Sinh is bijective. (Tip: Injectivity follows from Sinh being strictly positively increasing (which I have already proven in a different part) and use the intermediate value theorem to prove the surjectivity.)

My answer:

Due to $sinh(-x) = -sinh(x)$ it's enough to consider the interval $[0, \infty[$.

Injectivity:

Let $sinh$ be a non injective function, then there must exist $x, y \in \mathbb{R}$ such that $x > y$ with $sinh(x) = sinh(y)$. Due to sinh being strictly positively increasing (which I had already proven in a different part) this is a contradiction, so $x$ must equal $y$ and therefore the function is injective.

Surjectivity:

$f(x) := sinh(x)$

$f(0) = sinh(0) = \frac{e^0-e^{-0}}{2} = 0$

Due to the function being injective and strictly positively increasing, there must exist a $b \in \mathbb{R}$ such that $0 < b$ and $f(0)<f(b)$.

Due to the function being totally differentiable (Which I had already proven in another question), it is also continuous for all $x$. Out of this we can understand per the intermediate value theorem that for any $y \in \mathbb{R}$ such that $f(0)<y<f(b)$, there must exist an $x \in [0, b]$ such that $f(x) = y$. Now due to the continuity of the function, we can let $b$ tend to infinity and our requirement for the IVT will remain fulfilled, which means that the function is surjective and in turn bijective.

Now for the other part, I had to show the same for $tanh$, however I onlz had to show surjectivity for the intervall $]-1, 1[$.

Injectivity: Same as above.

Surjectivity: I first had to show that $|tanh(x)| < 1$ for all $x \in \mathbb{R}$ which I did like this:

$tanh(x) = \frac{sinh(x)}{cosh(x)} = \frac{e^x-e^{-x}}{e^x+e^{-x}}$

Now I just had to show that the numerator is always smaller than the denominator, which is clear, since zou basically have the same values, which are all larger than 0 (Its given at the beginning of the part and I dont need to prove it) but in the numerator you're subtracting two positive value, while in the denominator you're adding those same two positive value, which in turn means that it'll always be a fraction smaller than 1, and due to the numerator being the $sinh$ function and due to $sinh(-x) = -sinh(x)$ it follows then that it is also always larger than $-1$.

Now I just argued using the IVT again, saying since $-1 < y = tanh(x) < 1$ there must exist and $x \in ]-1,1[$ such that $f(x) = y$.

That was basically my answer. I have a feeling that I have made some logical mistakes, especially for the $tanh$ part and was hoping you guys would be able to let me know if what I did is correct and what I can improve.

## 1 Answer

"which in turn means that it'll always be a fraction smaller than 1, and due to the numerator being the sinhsinh function and due to sinh(−x)=−sinh(x) it follows then that it is also always larger than −1"

The numerator is 2sinh(x), but your argument still holds because $tanh(-x) = \frac{e^{-x} - e^x}{e^{-x} + e^x} = -tanh(x)$.

Except that small part, your proof seems good to me.

Edit: you have proved that the range of tanh(x) is not larger than ]-1, 1[, but you also have to show that for every $y \in ]-1, 1[$ there exists an $x$ such that $tanh(x) = y$, otherwise it is possible that the range of $tanh(x)$ is smaller than ]-1, 1[.

A hint for proving this is: $tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{e^{2x} - 1}{e^{2x} + 1}$

• Thanks for your answer. Would you be willing to give me another hint? Still stuch on that hint ;) – Rudy Ailabouni Jun 30 '17 at 13:38
• Let $y = e^{2x}$. Can you show that for every $z \in ]-1, 1[$, there exists a $y > 0$ such that $\frac{y - 1}{y + 1} = z$? – Pel de Pinda Jun 30 '17 at 13:42
• Would it be sufficient if I would let y tend to $\infty$ which means in turn that x tends to infinity since e^x tends to infinity for infinite x, and then do the same where I let y tend to 0? (Due to the point symmetry of the tanh function I dont really care about the negative interval) That would then mean that the larger the y, the closer it gets to one, and the smaller it is, the closer it gets to 0 and then due to the continuity of the function, and the monotony, I could argue that there must exist an x for each y using the IVT. – Rudy Ailabouni Jun 30 '17 at 13:54
• That would be the idea, but I think it is clearer to actually solve for y in the equation $\frac{y - 1}{y + 1} = z$, show that this solution exists and is positive, and then solve for $x$ in $y = e^{2x}$. – Pel de Pinda Jun 30 '17 at 13:56
• So my $y = \frac{z+1}{1-z}$ and $x = \frac{ln(z+1) - ln(1-z)}{2}$ y is always positive since $z \in \mathbb{R}$ and the $x$ exists for all z, since the only time $ln$ is undefined is when $z = 1/-1$ which is not in the interval where $z$ is defined. Is that correct? I have basically shown that for each $-1<z<1$ there exists an $x$ which fulfills $tanh$. Right? – Rudy Ailabouni Jun 30 '17 at 14:17