Verify each of the following by using equations (11.4), (12.2), and (12.3).

$\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathbf{3}\mathit{z}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{3}}\mathbf{\left(}\mathit{z}\mathbf{\right)}\mathbf{-}\mathbf{3}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathit{z}\mathbf{\right)}$

Given equation is$cos\left(3z\right)=4co{s}^3\left(z\right)-3cos\left(z\right)$.

The term "Hyperbolic Function" refers to the relationship between a point on a hyperbola's distance from its origin and its coordinate axes, expressed as a function of an angle.

Given the equationis$cos\left(3z\right)=4co{s}^3\left(z\right)-3cos\left(z\right)$.

The exponential form of the given equation is,

$cos\left(3z\right)=\frac{e^{\left(3zi\right)}+e^{\left(-3zi\right)}}{2}$ ….(1)

Let x(z),y(z) be $e^{\left(zi\right)}and{e}^{\left(-zi\right)}$ respectively.

Substitute values in equation (1).

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

Add $\left(2xy-2xy\right)$ in numerator.

$$\begin{gathered} cos\left(3z\right)=\frac{\left(x+y\right)\left(x^2-xy+y^2+2xy-2xy\right)}{2} \\ =\frac{\left(x+y\right)\left(x^2-2xy+y^2-3xy\right)}{2} \\ =\frac{\left(x+y\right)\left[\left(x+y{)}^2-3xy\right)\right]}{2} \\ =\frac{{\left(x+y\right)}^3-3xy\left(x+y\right)}{2} \end{gathered}$$

Replace the value of and .

$$\begin{gathered} cos\left(3z\right)=\frac{{\left[exp\left(z\right)+exp\left(-z\right)\right]}^3}{2}-\frac{3exp\left(z\right)+exp\left(-z\right)\left[exp\left(z\right)+exp\left(-z\right)\right]}{2} \\ =\frac{4{\left[exp\left(z\right)+exp\left(-z\right)\right]}^3}{8}-\frac{3exp\left(z\right)+exp\left(-z\right)}{2} \\ =4{\left(\frac{exp\left(z\right)+exp\left(-z\right)}{2}\right)}^3-\frac{3exp\left(z\right)+exp\left(-z\right)}{2} \\ =4co{s}^3\left(z\right)-3cos\left(z\right) \end{gathered}$$

Hence, the equation is verified.