Question

To find the values of the other hyperbolic functions at \(x\).

Short answer

The hyperbolic function are \(\sinh x = \frac{4}{3},{\mathop{\rm csch}\nolimits} x = \frac{3}{4},\tanh x = \frac{4}{5},\coth x = \frac{5}{4}\) and \(\sec hx = \frac{3}{5}\)

Step-by-step solution

Given function

The value \(\cos hx = \frac{5}{3}\) and \(x > 0\).

The concept of the hyperbolic function

(1) The hyperbolic sine function:\({sinhx = }\frac{{{1{e}^ 1{x}} 1{ - }{1{e}^{1{ - x}}}}}{1{2}}\)

(2) The hyperbolic cosine function:\({coshx = }\frac{{{{e}^{x}}{ + }{{e}^{{ - x}}}}}{{2}}\)

(3) The hyperbolic tangent function:\({tanhx = }\frac{{{sinhx}}}{{{coshx}}}\)

Use the formulas and find the hyperbolic function

The identity is \({\cosh ^2}x - {\sinh ^2}x = 1\).

\(\begin{aligned}{\cosh ^2}x - {\sinh ^2}x &= 1 \hfill \\{\sinh ^2}x &= {\cosh ^2}x - 1 \hfill \\|\sinh x| &= \sqrt {{{\cosh }^2}x - 1} \hfill \\|\sinh x| &= \sqrt {{{\sinh }^2}x} \hfill \\\end{aligned} \)

So, \(|\sinh x| = \sqrt {{{\cosh }^2}x - 1} \).

Since, \({|\rm{x| = x}}\) if \({\rm{x}} > {\rm{0}}\).

Substitute, \(\cosh x = \frac{5}{3}\).

\(\begin{aligned}{c}\sinh x &= \sqrt {{{\left( {\frac{5}{3}} \right)}^2} - 1} \\ &= \sqrt {\frac{{25}}{9} - 1} \\ &= \sqrt {\frac{{16}}{9}} \\ &= \frac{4}{3}\end{aligned}\)

Thus, the hyperbolic sine function is, \(\sinh x = \frac{4}{3}\).

The definition of the hyperbolic \({\mathop{\rm cosec}\nolimits} \)function is \({\mathop{\rm csch}\nolimits} x = \frac{1}{{\sin hx}}\).

The definition of the hyperbolic cosec function is \({\mathop{\rm csch}\nolimits} x = \frac{1}{{\sin hx}}\).

\({\mathop{\rm csch}\nolimits} x = \frac{1}{{\frac{4}{3}}}\quad \left( {{\rm{Q}}\sinh x = \frac{4}{3}} \right)\)

Thus, \({\mathop{\rm csch}\nolimits} x = \frac{3}{4}\).

Use the formula and find the function

The definition of the hyperbolic cosec function is \(\sec hx = \frac{1}{{\cos hx}}\).

\(\begin{array}{c}\sec hx = \frac{1}{{\frac{2}{3}}}\quad \\{\rm{Q}}\cos hx = \frac{5}{3}\end{array}\)

Thus, sec \(hx = \frac{3}{5}\).

The definition of the hyperbolic tangent function is \(\tanh x = \frac{{\sinh x}}{{\cosh x}}\).

\(\begin{array}{c}\tanh x = \frac{{\frac{4}{3}}}{{\frac{3}{3}}}\quad \\{\rm{Q}}\sinh x = \frac{4}{3}\\\cos hx = \frac{5}{3}\end{array}\)

Thus, \(\tanh x = \frac{4}{5}\).

The definition of the hyperbolic cot function is \(\cot hx = \frac{1}{{\sin hx}}\).

\(\begin{array}{c}\cot hx = \frac{1}{{\frac{4}{3}}}\quad \\{\rm{Q}}\tanh x = \frac{4}{5}\end{array}\)

Thus, \(\cot hx = \frac{5}{4}\).

Therefore, the hyperbolic function are \(\sinh x = \frac{4}{3},{\mathop{\rm csch}\nolimits} x = \frac{3}{4},\tanh x = \frac{4}{5},\coth x = \frac{5}{4}\) and \(\sec hx = \frac{3}{5}\).