Question

Verify the formulas in Problems 17 to 24.

${\mathbf{cosh}}^{\mathbf{-}\mathbf{1}}\mathbf{z}\mathbf{=}\mathbf{In}\left(z\pm \sqrt{z^2-1}\right)\mathbf{=}\mathbf{\pm }\mathbf{In}\left(z+\sqrt{z^2-1}\right)$

Short answer

The formula ${\cosh }^{-1}\mathrm{z}=\mathrm{In}(\mathrm{z}\pm \sqrt{{\mathrm{z}}^2-1})=\pm \mathrm{In}(\mathrm{z}+\sqrt{{\mathrm{z}}^2-1})$is verified.

Step-by-step solution

Given Information

Given formula is ${\cosh }^{-1}\mathrm{z}=\mathrm{In}(\mathrm{z}\pm \sqrt{{\mathrm{z}}^2-1})=\pm \mathrm{In}(\mathrm{z}+\sqrt{{\mathrm{z}}^2-1})$

Definition of Trigonometric equation.

A trigonometric equation is one that has one or more trigonometric ratios with unknown angles.

 Use exponential form to expand the equation

Given the function ${\cosh }^{-1}\mathrm{z}=\mathrm{In}(\mathrm{z}\pm \sqrt{{\mathrm{z}}^2-1})=\pm \mathrm{In}(\mathrm{z}+\sqrt{{\mathrm{z}}^2-1}).$

Write the exponential form of the z=cosh(w).

$$\begin{gathered} z=\frac{e^{\left(w\right)}+e^{\left(-w\right)}}{2} \\ {e}^{\left(2w\right)}-2e^{\left(w\right)}+1=0\dots \left(1\right) \end{gathered}$$

Let$u=e^{\left(w\right)}$ in equation (2).

${\mathrm{u}}^2-2\mathrm{zu}+1=0\dots \left(2\right)$

The coefficient of equation is as follows.

a=1

b=-2z

c=1

Use quadratic formula to find roots of equation (2).

$$\begin{gathered} u=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ =\frac{2z\pm \sqrt{{\left(-2z\right)}^2-4}}{2} \\ =z\pm \sqrt{z^2-1} \end{gathered}$$

Replace value of$ubyu=e^{\left(w\right)}$.

$w=In\left(z\pm \sqrt{z^2-1}\right)$

Replace value of$wbyw=\cos h^{-1}\left(z\right).$

$\cos h^{-1}z=In\left(z\pm \sqrt{z^2-1}\right)$

Prove cosh-1z=±In(z+z2-1) to prove the formula.

Take z=cosh(w) and if cosh(w) is even function than cosh(-w)=cosh(w)

Therefore, the formula is as follows.

$$\begin{gathered} -w=\cos h^{-1}\left(z\right) \\ w=\cos h^{-1}\left(z\right) \end{gathered}$$

The above condition satisfy$\pm w=In\left(z\pm \sqrt{z^2-1}\right)$ which proves the formula.

$w=\pm In\left(z\pm \sqrt{z^2-1}\right)$

Replace value of$wbyw=\cos h^{-1}\left(z\right)$.

role="math" localid="1658902884147" $\cos h^{-1}z=In\left(z\pm \sqrt{z^2-1}\right)......\left(3\right)$

Replace value of $wbyw=-\cos h^{-1}\left(z\right)$

role="math" localid="1658902896732" $\cos h^{-1}z=-In\left(z\pm \sqrt{z^2-1}\right)\left(4\right)$

Combine equations (3) and (4).

$\cos h^{-1}z=\pm In\left(z+\sqrt{z^2-1}\right)$

Hence, the formula is verified.