If $x=\sinh{\theta}$, is it possible to express $\cosh{n\theta}$ and $\sinh{n\theta}$ in terms of $x$?

Question

We know that hyperbolic sine is: $$\sinh \theta={\frac {e^{\theta}-e^{-\theta}}{2}}$$ and that hyperbolic cosine is $$\cosh \theta={\frac {e^{\theta}+e^{-\theta}}{2}}$$ Let $n\in\mathbb N$.

If $x=\sinh{\theta}$, is it possible to express $\cosh{n\theta}$ and $\sinh{n\theta}$ in terms of $x$?

Answer 1

Hint

May be, you could try to use $$\sinh(x)=\sin(ix) \qquad \text{and} \qquad \cosh(x)=\cos(ix)$$

Answer 2

These hyperbolic identities, very similar to the trigonometric ones, allow writing $\sinh nt$ and $\cosh nt$ (I use $t$ because the parameter does not strictly correspond to an angle in the geometric sense) in terms of $\sinh x=q$: $$\sinh(x+y)=\sinh x\cosh y+\cosh x\sinh y$$ $$\cosh(x+y)=\cosh x\cosh y+\sinh x\sinh y$$ $$\cosh^2x-\sinh^2x=1\implies\cosh x=\sqrt{1+q^2}$$

Answer 3

$\begin{align}\sinh(\theta)=\frac{e^\theta-e^{-\theta}}{2}=x &\Rightarrow e^{2\theta}-2xe^{\theta}-1 =0\\ &\Rightarrow e^{\theta}= \frac{2x\pm\sqrt{4x^2+4}}{2} = x\pm\sqrt{x^2+1} \\ &\Rightarrow \boxed{e^{\theta} = x+\sqrt{x^2+1}} ~~~\text{ as } e^\theta \ge0,\theta \in \Bbb R\end{align}$

So, $$\sinh(n\theta) = \frac{e^{n\theta}-e^{-n\theta}}{2} = \frac{(x+\sqrt{x^2+1})^n-(x+\sqrt{x^2+1})^{-n}}{2}$$

and

$$\cosh(n\theta) = \frac{e^{n\theta}+e^{-n\theta}}{2} = \frac{(x+\sqrt{x^2+1})^n+(x+\sqrt{x^2+1})^{-n}}{2}$$

Answer 4

Hint:

• You can express $\sin nt$ and $\cos nt$ in terms of Chebyshev polynomials of $\sin t$ and $\cos t$

• Hyperbolic and circular functions are related by $\sin it = i\sinh t$ and $\cos it = \cosh t$

• $\cosh t$ and $\sinh t$ are related by $\cosh^2t - \sinh^2 t = 1$

For the first hint, see the page https://mathworld.wolfram.com/Multiple-AngleFormulas.html