Question

If $\sinh x = \dfrac{3}{2}$, how do you find exact values of $\cosh x$ and $\tanh x$?

Answer 1

Hint: We will use some standard trigonometric identities and formulas to find the exact values of $\cosh x$ and $\tanh x$. To find $\cosh x$ we will use the relationship between $\sinh $ and $\cosh $. And to find $\tanh x$ we take the ratio of $\sinh $ and $\cosh $.

Complete step by step answer:
We know that,
\[{\cosh ^2}x - {\sinh ^2}x = 1\]
Substituting the value of $\sinh x = \dfrac{3}{2}$,
\[{\cosh ^2}x - {\left( {\dfrac{3}{2}} \right)^2} = 1\]
Taking $\dfrac{3}{2}$ to the right-hand side,
\[{\cosh ^2}x = 1 + {\left( {\dfrac{3}{2}} \right)^2}\]
Taking square of $\dfrac{3}{2}$,
\[{\cosh ^2}x = 1 + \dfrac{9}{4}\]
Taking L.C.M on right-hand side,
\[{\cosh ^2}x = \dfrac{{4 + 9}}{4}\]
Adding the terms,
\[{\cosh ^2}x = \dfrac{{13}}{4}\]
Taking square root on both sides,
\[\cosh x = \sqrt {\dfrac{{13}}{4}} = \dfrac{{\sqrt {13} }}{2}\]

Now, we know,
$\tanh x = \dfrac{{\sinh x}}{{\cosh x}}$
Substituting the values of $\sinh $ and $\cosh $,
$\tanh x = \dfrac{{\dfrac{3}{2}}}{{\dfrac{{\sqrt {13} }}{2}}}$
Simplifying the right-hand side,
$\tanh x = \dfrac{3}{{\sqrt {13} }}$

Therefore, \[\cosh x = \dfrac{{\sqrt {13} }}{2}\] and $\tanh x = \dfrac{3}{{\sqrt {13} }}$.

Note:
For hyperbolic trigonometric identities we use \[{\cosh ^2}x - {\sinh ^2}x = 1\] whereas for normal trigonometric identities we use \[{\cos ^2}x + {\sin ^2}x = 1\]. While solving such types of problems, students may get confused with all the trigonometric formulas, identities and properties.