How would you write $\sinh(4θ)$ in terms of $\cosh(4θ)$ and $\sinh(4θ)$, all in relation to the function $e^{4θ}$

Question

I know the hyperbolic identites but I cant seem to find a way to do it involving $\cosh(4θ)$ and $\sinh(4θ)$.

Answer 1

I think you mean writing $\cosh(4\theta)$ and $\sinh(4\theta)$ in terms of $e^{4\theta}$ in which case,
$$\cosh(4\theta)=\frac{e^{4\theta}+e^{-4\theta}}{2}$$
$$\sinh(4\theta)=\frac{e^{4\theta}-e^{-4\theta}}{2}$$
Using the above, you get : $$\cosh^2 (4\theta)-\sinh^2 (4\theta)=1$$