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

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 $$x$$?

Hint

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

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 $$\sinh x=q$$: $$\sinh(x+y)=\sinh x\cosh y+\cosh x\sinh y$$ $$\cosh(x+y)=\cosh x\cosh y+\sinh x\sinh y$$ $$\cosh^2x-\sinh^2x=1\implies\cosh x=\sqrt{1+q^2}$$

\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

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

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