Show that ${\mathbf{e}}^{\mathbf{nz}}\mathbf{=}\mathbf{(}\mathbf{coshz}\mathbf{+}\mathbf{sinhz}{\mathbf{)}}^{\mathbf{n}}\mathbf{=}\mathbf{cosh}\mathbf{}\mathbf{nz}\mathbf{+}\mathbf{sinh}\mathbf{}\mathbf{nz}\mathbf{.}$Use this and a similar equation for ${\mathit{e}}^{\mathbf{-}\mathbf{n}\mathbf{z}}$to find formulas for $\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathbf{3}\mathit{z}$and $\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{3}\mathit{z}$in terms of $\mathbf{sinhz}$ and $\mathbf{coshz}$.

The equation to prove is,$e^{n}z=\cosh nz+\sinh nz$ .

A hyperbolic function is a representation of the relationship between a point's distances from the origin to the coordinate axes as a function of an angle.

The equation to prove is,

$e^{nz}=\cosh \mathrm{nz}+\sinh \mathrm{nz}$ …(1)

Take Left Hand Side of the equation (1) and replace $nz==-\left(inz\right)i$.

$$\begin{gathered} e^{nz}=e^{\left[-\left(inz\right)i\right]} \\ =\cos \left(nzi\right)-i\sin \left(nzi\right) \end{gathered}$$ …(2)

Use the following identities in equation (2).

$$\begin{gathered} \cos \left(nzi\right)=\cos h\left(nz\right) \\ \sin \left(nzi\right)=i\sin h\left(nz\right) \end{gathered}$$

Hence the required result isrole="math" localid="1658811404773" $e^{nz}=cosh\left(nz\right)+sinh\left(nz\right)$ . …. (3)

The equation to prove is$e^{nz}=cosh\left(nz\right)+sinh\left(nz\right)$ …. (4)

Take Left Hand Side of the equation (1) and replace $(-nz)=\left(inz\right)i$.

$$\begin{gathered} e^{\left(-nz\right)}=e^{\left(inz\right)i} \\ =\cos \left(nzi\right)+i\sin \left(nzi\right) \end{gathered}$$ …(5)

Use the following identities in equation (5).

$$\begin{gathered} \cos h\left(nzi\right)=\cos h\left(nz\right) \\ \sin \left(nzi\right)=i\sin h\left(nz\right) \end{gathered}$$

Hence the required result is $e^{-nz}=cosh\left(nz\right)-sinh\left(nz\right)$ …(6).

The exponential form of $cosh\left(3z\right)=\frac{e^{3z}+e^{-3z}}{2}$. …. (7)

Put $(e^z,e^{-z})=(x,y)$in equation (7) …. (8)

role="math" localid="1658812377265" $$\begin{gathered} \cos h3z=\frac{x^3+y^3}{2} \\ =\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{2} \\ =\frac{\left(x+y\right)\left(x^2-xy+y^2+2xy+-2xy\right)}{2} \\ =\frac{\left(x+y\right)\left(x^2+2xy+y^2-3xy\right)}{2} \end{gathered}$$

Replace $x^2+2xy+y^2$by $(x+y{)}^2$.

role="math" localid="1658812458369" $$\begin{gathered} \cos h3z=\frac{\left(x+y\right)\left[{\left(x+y\right)}^2-3xy\right]}{2} \\ =\frac{\left[{\left(x+y\right)}^3-3xy\left(x+y\right)\right]}{2} \\ =\frac{4}{4}\times \frac{{\left(x+y\right)}^3}{2}-\frac{3xy\left(x+y\right)}{2} \\ =\frac{4{\left(x+y\right)}^3}{8}-\frac{3xy\left(x+y\right)}{2} \end{gathered}$$

Use equation (8) in the above equation.

$$\begin{gathered} \cos h3z=4\left(\frac{e^z+e^{-z}}{2}\right)-\frac{3\left[e^z+e^{-z}\right]}{2} \\ =4\cos h^3\left(z\right)-3\cos h\left(z\right) \end{gathered}$$

Hence the formula of $cosh3x=4cos{h}^3\left(z\right)-3cosh\left(z\right)$

The exponential form of $sinh\left(3z\right)=\frac{(e^{3}z-e^{-3z})}{2}$. …. (9)

Put $(e^z,e^{-z})=(x,y)$in equation (9) …. (10)

$$\begin{gathered} \sin h3z=\frac{x^3-y^3}{2} \\ =\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{2} \\ =\frac{\left(x-y\right)\left(x^2+xy+y^2+2xy+-2xy\right)}{2} \\ =\frac{\left(x-y\right)\left(x^2-2xy+y^2+3xy\right)}{2} \end{gathered}$$

Replace $x^2-2xy+y^2$by $(x-y{)}^2$.

$$\begin{gathered} \sin h3z=\frac{\left(x-y\right)\left[{\left(x-y\right)}^2+3xy\right]}{2} \\ =\frac{\left[{\left(x-y\right)}^3+3xy\left(x-y\right)\right]}{2} \\ =\frac{4}{4}\times \frac{{\left(x-y\right)}^3}{2}-\frac{3xy\left(x-y\right)}{2} \\ =\frac{4{\left(x-y\right)}^3}{8}-\frac{3xy\left(x-y\right)}{2} \end{gathered}$$

Use equation (10) in the above equation.

$$\begin{gathered} \sin h3z=4\left(\frac{e^z-e^{-z}}{2}\right)+\frac{3\left[e^z-e^{-z}\right]}{2} \\ =4\sin h^3\left(z\right)+3\sin h\left(z\right) \end{gathered}$$

Hence the formula of $sinh3x=4sin{h}^3\left(z\right)+3sinh\left(z\right)$